MATHEMATICS FOR ENGINEERS
PART II
The Directly-Useful Technical Series
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Mathematics for Engineers
By W. N. ROSE, B.Sc. Eng. (Lond.)
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To the young engineer it will be a God-send." — Managing Engineer.
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*
The Directly-Useful
Technical Series
FOUNDED BY THE LATE WILFRID J. LINEHAM, B.Sc., M.Inst.C.E.
Mathematics for Engineers
PART II
BY
W. N. ROSE
B.Sc. ENG. (LOND.)
Late Lecturer in Engineering Mathematics at the
University of London Goldsmiths' College
Teacher of Mathematics, Borough
Polytechnic Institute
LONDON
CHAPMAN & HALL, LTD.
11 HENRIETTA STREET, W.C. 2
1920
PRINTED IN GREAT BRITAIN BY
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EDITORIAL NOTE
THE DIRECTLY-USEFUL TECHNICAL SERIES requires a few words
by way of introduction. Technical books of the past have arranged
themselves largely under two sections : the Theoretical and the
Practical. Theoretical books have been written more for the
training of- college students than for the supply of information to
men in practice, and have been greatly filled with problems of an
academic character. Practical books have often sought the other
extreme, omitting the scientific basis upon which all good practice
is built, whether discernible or not. The present series is intended
to occupy a midway position. ,The information, the problems and
the exercises are to be of a directly-useful character, but must at
the same time be wedded to that proper amount of scientific
explanation which alone will satisfy the inquiring mind. We
shall thus appeal to all technical people throughout the land, either
students or those in actual practice.
AUTHOR'S PREFACE
CONCERNING the aim and scope of this work nothing need be added
here to the statement made in the Preface to the former volume..
It is there asserted that the subject-matter has been so chosen, and,
through the examples, so applied to practical problems, that the
two volumes " embrace all the mathematical work needed by-
engineers in their practice, and by students in all branches of
engineering science."
As with the first volume, much thought has been given to the
elimination of all rules and processes of academic interest only ;
but the fact of the importance and necessity of logical reasoning
has not been overlooked.
With the exception of the chapters on Spherical Trigonometry
and Mathematical Probability, this volume is devoted to the study
of the Calculus, both Differential and Integral. Whilst it is wise,
and even imperative, if this subject is to be presented in an intelligible
manner, that much attention should be paid to the graphic interpre-
tation of its rules, care must be taken to ensure that the graphic
methods do not become other than auxiliaries. Accordingly the
treatment throughout is based upon algebraic principles ; but
whenever graphic proofs or constructions have been found to amplify
or explain the subject, they have been utilised to the fullest extent.
Thus from the commencement the connection between the rate of
change of a quantity and the slope of a curve is clearly demonstrated :
and this correlation of the algebraic and the graphic methods is con-
tinued through all the stages of the development of the subject.
The conception of " limiting values," mentioned briefly in Part I,
is further discussed in Chapter I, a familiar example from Dynamics
being chosen as the illustration : in this chapter also two methods
of graphic differentiation are given ; the second of which, and the
less familiar, being the simpler to apply.
viii AUTHOR'S PREFACE
The various rules for the differentiation of both algebraic and
trigonometric functions are explained in detail in Chapter II ;
and Chapter III, containing the rules for the differentiation of a
function of a function, a product of functions, etc., together with an
introduction to partial differentiation, may be regarded as comple-
mentary to Chapter II.
From the abstract reasoning required for comprehension of such
an idea as that of " limiting values/' the practical mind turns with
relief to the applications of differentiation found in Chapter IV ;
the determination of maximum and minimum values making a
particularly strong appeal. In view of the importance of this
branch of the subject, a very varied selection of practical examples
is presented, in the choice of which the method of solution has
been a determining factor. In this chapter also the use of Taylor's
theorem in cases of interpolation from steam tables is demonstrated.
Chapters V and VI contain the rules required for the integration
of functions occurring in engineering theory and practice. The
former chapter serves as an introduction to integration, the signifi-
cance of the symbols f and dx being explained by reference to a
graph ; whilst in the latter chapter the various types of integrals,
many of them of a somewhat complicated character, are discussed.
At this stage also the reduction formulae are introduced, and mention
is made of the Gamma function and its uses.
Instances of the application of the rules of integration are to be
found in the processes enumerated in Chapter VII ; and special
features of this chapter are the determination of the perimeter of
the ellipse, the graphic method for fixing the position of the centroid
vertical, the drawing of ist and 2nd moment curves and the evalua-
tion of the moment of inertia of a compound vibrator.
The utility of polar co-ordinates to the electrical engineer is shown
by the inclusion of examples on the candle-powers of lamps, and the
employment of the Rousseau diagram to find the mean spherical
candle-power ; and Dr. Fleming's graphic method for determina-
tion of root mean square values of currents is here inserted, since
it involves polar plotting.
Differential equations occur so frequently that the methods of
solution demand most careful study. Chapter IX presents the most
common types, and the selection of examples based upon these,
both worked and set, emphasises the need for a proper appreciation
of the method of solution.
Chapter X, with its applications of the Calculus to problems
encountered in the study of Thermodynamics, Strength of Materials,
AUTHOR'S PREFACE ix
Applied Mechanics, Applied Electricity and Hydraulics, provides
further illustration of the need of a sound knowledge of the subject
to the engineer desirous of equipping himself at all points.
The last two chapters contain much of interest to the surveyor,
the examples chosen being such as arise in his practice ; and par-
ticular attention is directed to the investigation relating to the
corrections following errors of observation.
The Author greatly deplores the fact that the inspirer of this work,
the late Mr. W. J. LINEHAM, B.Sc., M.I.C.E., does not see its com-
pletion : to his enthusiasm for his ideals in education, and for his
many personal kindnesses to the Author, tribute is here paid.
Sincere thanks are also tendered to Messrs. J. L. BALE and C. B.
CLAPHAM, B.Sc., for much valuable assistance.
Great care has been taken to produce the book free from errors,
but some may remain, notification of which will be esteemed a great
favour.
W. N. ROSE.
Borough Polytechnic Institute,
S.E. i,
December, 1919.
CONTENTS
PAGE
INTRODUCTORY i
Abbreviations.
CHAPTER *I
INTRODUCTION TO DIFFERENTIATION 3
Historical note — Rates of change — Average and actual rates of
change — Slopes of curves — Graphic differentiation by two methods.
CHAPTER II
DIFFERENTIATION OF FUNCTIONS 26
Differentiation of ax" — Differentiation of a sum of terms — Proof
of construction for slope curves — Beam problems — Lengths of
sub-tangents and sub-normals of curves — Differentiation of ex-
ponential functions — Differentiation of log* — Differentiation of
sinh x and cosh x — Differentiation of the trigonometric functions —
Simple harmonic motion.
CHAPTER III
ADDITIONAL RULES OF DIFFERENTIATION .... 63
Differentiation of a function of a function — Differentiation of a
product — Differentiation of a quotient — Differentiation of inverse
trigonometric functions — Partial differentiation — Total differential
—Logarithmic differentiation.
CHAPTER IV
APPLICATIONS OF DIFFERENTIATION ...... 88
Maximum and minimum values — Point of inflexion — Calculation
of small corrections — Expansion of functions in series — Theorems
of Taylor and Maclaurin.
xii CONTENTS
CHAPTER V
PAGE
INTEGRATION 115
Meaning of integration — Graphic integration — Application of in-
tegration to " beam " problems — Coradi integraph — Rules for
integration of simpler functions — Integration of powers of x —
Integration of exponential functions— Integration of trigonometric
functions — Indefinite and definite integrals — Method of determining
the values of definite integrals — Proof of Simpson's rule.
CHAPTER VI
FURTHER METHODS OF INTEGRATION . . . . . . 146
Integration by the aid of partial fractions — Integration by the
resolution of a product into a sum — Integration by substitution —
Integration by parts — Reduction formulae — Gamma function —
List of integrals.
CHAPTER VII
MEAN VALUES, ETC. . . • . . . . . . 180
Determination of mean values — Root mean square values —
Volumes — Volumes of solids of revolution — Length of arc — Peri-
meter of ellipse — Area of surface of solid of revolution— Centre of
gravity — Centroid — " Double sum curve " method of finding the
centroid vertical — Centroids of sections by calculation — Centroids
found by algebraic integration — Centre of gravity of irregular
solids — Centre of gravity of a solid of revolution — Centre of pressure
— Moment of inertia — Swing radius — The parallel axis theorem —
Theorems of perpendicular axes — Moment of inertia of compound
vibrators — Determination of ist and 2nd moments of sections by
means of a graphic construction and the use of a planimeter.
CHAPTER VIII
POLAR CO-ORDINATES 257
Polar co-ordinates — Spirals — Connection between rectangular and
polar co-ordinates — Use of polar co-ordinates for the determination
of areas — The Rousseau diagram — Dr. Fleming's graphic method for
the determination of R.M.S. values — Theory of the Amsler plani-
meter.
CONTENTS xiii
CHAPTER IX
PAGE
SIMPLE DIFFERENTIAL EQUATIONS . - 270
Differential equations, definition and classification — Types : ~-
dy
given as a function of x : -~ given as a function of y : General linear
equations of the first order : Exact differential equations : Equa-
tions homogeneous in x and y : Linear equations of the second order
— Use of the operator D — Useful theorems involving the operator D
— Equations of the second degree.
CHAPTER X
APPLICATIONS OF THE CALCULUS 300
Examples in Thermodynamics : Work done in the expansion of a
gas — Work done in a complete theoretical cycle — Entropy of water
- — Efficiency of engine working on the Rankine cycle — Efficiency of
engine working on the Rankine cycle, with steam kept saturated by
jacket steam — Examples relating to loaded beams, simply supported
or with fixed ends, the loading and the section varying — Shearing
stress in beams — Examples on Applied Electricity — Examples on
Strengths of Materials — Loaded struts — Tension in belt — Friction
in a footstep bearing — Schiele pivot — Examples on Hydraulics —
Centre of Pressure — Transition curve in surveying.
CHAPTER XI
HARMONIC ANALYSIS . .- . 342
Fourier's theorem — Analysis by calculation — Harrison's graphic
method of analysis — Analysis by superposition.
CHAPTER XII
THE SOLUTION OF SPHERICAL TRIANGLES . . . • 355
Definition of terms — Spherical triangle — Solution of spherical
triangles — Solution of right-angled spherical triangles — Napier's
rules of circular parts — The "ambiguous" case — Applications in
spherical astronomy — Graphic solution of a spherical triangle.
CHAPTER XIII
MATHEMATICAL PROBABILITY AND THEOREM OF LEAST
SQUARES 370
Probability — Exclusive events — Probability of the happening
together of two independent events — Probability of error —
Theorem of least squares — Error of the arithmetic mean — Weight
of an observation.
xiv .CONTENTS
PAGE
ANSWERS TO EXERCISES . 387
TABLES :—
Trigonometrical ratios 397
Logarithms 398
Antilogarithms 400
Napierian logarithms 402
Natural sines 404
Natural cosines 406
Natural tangents 408
Logarithmic sines 410
Logarithmic cosines , .412
Logarithmic tangents . . . . . . . . . .414
Exponential and hyperbolic functions 416
INDEX ...... 417
MATHEMATICS FOR ENGINEERS
PART II
INTRODUCTORY
THE subject-matter of this volume presents greater difficulty
than that of Part I. Many of the processes described herein de-
pend upon rules explained and proved in the former volume ;
and accordingly it is suggested that, before commencing to read
this work, special attention should first be paid to Part I, pp. 452-
460, 463-467, 469-472, and pp. 273-299 ; whilst a knowledge of the
forms of the curves plotted in Chapter IX should certainly prove
of great assistance.
The abbreviations detailed below will be adopted throughout.
-> stands for " approaches."
„ " equals " or " is equal to."
+ ,, ,, " plus."
„ ,, " minus."
X „ „ " multiplied by."
4- „ „ " divided by."
„ „ " therefore."
± „ „ " plus or minus."
„ ,, " greater than."
" less than."
0 ,, „ " circle."
©ce ,, ,, " circumference."
« „ ,, " varies as."
co „ ,, " infinity."
L. ,, ,, " angle."
A >, ,, triangle " or " area of triangle."
Li_ or 4 ! „ „ " factorial four " ; the value being that of the
product 1.2.3.4 or 24-
"P, „ „ " the number of permutations of n things taken
two at a time."
"C2 „ „ " the number of combinations of n things taken
two at a time."
n.3 „ „ n (n — i) (n — 2).
B 1
2 MATHEMATICS FOR ENGINEERS
t\ stands for " efficiency."
0 „ „ " angle in degrees."
6 ,, ,, " angle in radians."
I.H.P. ,, „ " indicated horse-power."
B.H.P. ,, „ " brake horse-power."
m.p.h. „ ,, " miles per hour."
r.p.m. ,, ,, " revolutions per minute."
r.p.s. „ „ " revolutions per second."
I.V. ,, ,, " independent variable."
F.° ,, „ " degrees Fahrenheit."
C.° ., ,, " degrees Centigrade."
E.M.F. ,, ,, " electro-motive force."
1 ,, .., " moment of inertia."
E „ „ " Young's modulus of elasticity."
Sn „ ,, " the sum to n terms."
S^ ,, „ " the sum to infinity (of terms)."
2 ,, ,, " sum of."
B.T.U. „ „ " Board of Trade unit."
B.Th.U. „ „ " British thermal unit."
T ,, ,, " absolute temperature."
M ,, ,, " coefficient of friction."
sin"1 x ,, ,, " the angle whose sine is x."
e ,, ,, " the base of Napierian logarithms."
g ,, the acceleration due to the force of gravity."
cms. ,, ,, " centimetres."
grms. ,, ,, " grammes."
-Ly ,, „ " the limit to which y approaches as x approaches
the value a."
C. of G. ,, ,, " centre of gravity."
C. of P. „ ,, " centre of pressure."
k ,, ,, " swing radius," or " radius of gyration."
M.V. „ ,, " mean value."
R.M.S. „ „ " root mean square."
f'(x] ,, ,, " the first derivative of a function of x."
f"(x] ,, ,, " the second derivative of a function of x."
dy
-j- ,, ,, " the differential coefficient of y with regard to x."
I ydx ,, ,, " the integral of y with respect to x as the I.V."
8 ,,..," difference of."
,, " the operation -3-.."
" candle-power."
M.S.C.P. „ ,, " mean spherical candle-power."
p ,, ,, " density."
CHAPTER I
INTRODUCTION TO DIFFERENTIATION
THE seventeenth century will ever be remarkable for the number
of great mathematicians that it produced, and still more so for the
magnitude of the research accomplished by them. In the early
part of the century Napier and Briggs had introduced their systems
of logarithms/ whilst Wallis and others directed their thoughts to
the quadrature of curves, which they effected in some instances
by expansion into series, although the Binomial Theorem was then
unknown to them. In 1665 Newton, in his search for the method of
quadrature, evolved what he termed to be a system of " fluxions " or
flowing quantities : if x and y, say, were flowing quantities, then
he denoted the velocity by which each of these fluents increased by
x and y respectively. By the use of these new forms he was enabled
to determine expressions for the tangents of curves, and also for
their radii of curvature. At about the same time Leibnitz of Leipsic,
also concerned with the same problem, arrived at practically the
same system, although he obtained his tangents by determining
" differences of numbers." To Leibnitz is due the introduction of
the term " differential," and also the differential notation, viz.,
dx and dy for the differentials of x and y : he also in his expression
for the summation of a number of quantities first wrote the symbol
f, his first idea being to employ the word " omnia " or its abbre-
viation " omn~." Thus, if summing a number of quantities like x,
he first wrote " omnia x," which he contracted to " omn. x," and
later he modified this form to fx.
Great controversy raged for some time as to the claims of
Newton and Leibnitz to be called the inventor of the system of the
" Calculus," which is" the generic term for a classified collection of
rules; but it is now generally conceded that the discoveries were
independent, and were in fact the natural culmination of the
research and discoveries of many minds.
3
MATHEMATICS FOR ENGINEERS
The Calculus was further developed by Euler, Bernoulli, Legendre
and many others, but until a very recent date it remained merely
" a classified collection of rules" : its true meaning and the wide
field of its application were for long obscured.
Nowadays, however, a knowledge of the Calculus is regarded,
particularly by the engineer, as a vital part of his mental equip-
ment : its rules have been so modified as to become no serious tax
on the memory, and the true significance of the processes has
been presented in so clear a light that the study of the Calculus
presents few difficulties even to the ultra-practical engineer.
This revolution of thought has been brought about entirely
through the efforts of men who, realising the vast potentialities
of the Calculus, have reorganised the teaching of the subject :
they have clothed it and made it a live thing.
The Calculus may be divided into two sections, viz., those
treating of differentiation and integration respectively. Differentia-
tion, as the name suggests, is that part of the subject which is
concerned with differences, or more strictly with the comparison
of differences of two quantities. Thus the process of differentia-
tion resolves itself into a calculation of rates of change; but the
manner in which the rate of change is determined depends on the
form in which the problem is stated. Thus, if the given quantities
are expressed by the co-ordinates of a curve, the rate of change of
the ordinate compared with the change in the abscissa for any
particular value of the abscissa is measured by the slope of the
curve at the point considered.
Differentiation is really nothing more nor less than the deter-
mination of rates of change or of slopes of curves.
The term " rate of change " does not necessarily imply a " time
rate of change," i. e., a rate of change with regard to time, such as
the rate at which an electric current is changing per second, or the
rate at which energy is being stored per minute; but the change
in one quantity may be compared with the change in any other
quantity. As an illustration of this fact we may discuss the
following example —
The velocity of a moving body was measured at various distances
from its starting point and the results were tabulated, thus —
s (distance in feet) .
o
5
12
v (velocity in feet per sec.)
10
M
15
INTRODUCTION TO DIFFERENTIATION 5
To find the values of the " space rate of change of velocity " for
the separate space intervals.
Considering the displacement from o to 5 ft., the change in the
velocity corresponding to this change of position is 14—10, *. e.,
4 ft. per sec.
change of velocity 14—10 4
Hence — —* .^— = - - = - = -8
change of position 5 — o 5
or, the change of velocity per one foot change of position = -8 ft.
per sec., and rate of change of velocity = -8 ft. per sec. per foot.
Again, if s varies from o to 12, the change of v = 15—10 = 5 ;
or, the rate of change of velocity (for this period) = T5^ ft. per
sec. per foot.
Similarly, the rate of change of v, whilst s ranges from 5 to 12,
15—14 i ,, ,
— -^— - = - ft. per sec. per foot.
12- 5 7
The rates of change have thus been found by comparing differ-
ences. The phrase " change of" occurs frequently in this investi-
gation, and to avoid continually writing it a symbol is adopted in
its place. The letter thus introduced is 8 (delta), the Greek form
of d, the initial letter of the word " difference " : it must be regarded
on all occasions as an abbreviation, and hence no operation must
be performed upon it that could not be performed if the phrase for
which S stands was written in full. In other words, the ordinary
rules applying to algebraic quantities, such as multiplication,
division, addition or subtraction, would be incorrectly used in
conjunction with 8.
Thus, mv (the formula for momentum) means m multiplied by v,
or a mass multiplied by a velocity, whilst Sv represents " the change
of v," or if v is the symbol for velocity, 8v — change of velocity.
Again, 8/ = change of time or change of temperature, as the
case may be. Using this notation our previous statements can be
written in the shorter forms : thus—
(1) As s changes from o to 5 Sv = 14—10 = 4
8s = 5-0 = 5
H-*
(2) As s changes from o to 12 Sv = 15 — 10 = 5
8s = 12— o = 12
and !_y = A
Ss 12 * *
MATHEMATICS FOR ENGINEERS
(3) As s changes from 5 to 12
Sv i
and ~=s — s= •!<:
8v = 15 — 14 = i
Ss = 12 — 5 = 7
It must be noted that we do not cancel S from the numerator and
denominator of the fraction ^-.
The final result in (i), viz., ^- = -8, as s changes from o to 5, needs
further qualification. From the information supplied we cannot
say with truth that the change in the velocity for each foot from
o to 5 ft. is -8 ft. per sec. : all that we know with certainty is that,
as s changes from o to 5 ft., the average rate of change of velocity
over this space period is -8 ft. per sec. Supposing the change of
velocity to be continuous over the period considered, the value of ^
already obtained would be the actual rate of change of velocity at
some point or points in the period considered.
It is usual to tabulate the values of the original quantities and
their changes, and unless anything is given to the contrary the
average values of the rate of change are written in the middle of the
respective periods.
The table is set out thus —
Sv
s
V
Ss
Sv
Ss
0
IO
—
—
5
4
i = -8
5
14
—
—
—
7
i
r = '143
12
15
~~~"
~~
_—.
To distinguish in writing between average and actual rates of
change the notation employed is slightly modified, d being used in
dv
place of 8 ; -j- thus representing an actual rate of change of velocity,
and ^7 representing an average rate of change of velocity. Once
again it must be emphasised that d must be treated strictly in
association with the v or t, as the case may be, and dt does not mean
, , j dv • v
dxt, nor does -j- give .
INTRODUCTION TO DIFFERENTIATION
Another example can now be considered to demonstrate clearly
the distinction between an average and an actual rate of change.
For a body falling freely under the influence of gravity the
values of the distances covered to the ends of the ist, 2nd and 3rd
seconds of the motion are as in the table —
t (sees.) .
o
I
2
3
s (feet) .
0
16.1
64.4
144.9
Find the average velocities during the various intervals of time,
and also the actual velocities at the ends of the ist, 2nd and 3rd
seconds respectively.
The average velocities are found in the manner described before,
i. e., by the comparison of differences of space and time, and the
results are tabulated, thus —
'
'
5s
st
v-Ss
o
0
—
—
16-1
i
16-1
I
16-1
—
—
—
—
—
48-3
i
48-3
2
64-4
—
—
—
80-5
i
80-5
3
144-9
~~~"~
~
The average velocities, viz., the values in the last column, are
written in the lines between the values of the time to signify that
they are the averages for the particular intervals. As also it is
known that in this case the velocity is increased at a uniform rate,
it is perfectly correct to state that the actual velocities at the ends
of -5, 1-5 and 2-5 seconds respectively are given by the average
velocities over the three periods and are 16-1, 48-3 and 80-5 ft.
per sec.
We have thus found the actual velocities at the half seconds,
but not those at the ends of the ist, 2nd and 3rd seconds. The
determination of these velocities introduces a most important
process, illustrating well the elements of differentiation, and in
consequence the investigation is discussed in great detail. .
The student of Dynamics knows that -the law connecting space
and time, in the case of a falling body, is s = \gtz = i6-i/2, and
8 MATHEMATICS FOR ENGINEERS
a glance at the table of values of s and t confirms this law ; thus,
when t = 2, s = 64-4, which = i6-iX22 or i6-i^2.
To find the actual velocity at the end of the first second we
must calculate the average velocities over small intervals of time
in the neighbourhood of I sec., and see to what figure these velocities
approach as the interval of time is taken smaller and smaller.
Thus if — t — i s = 16-1 x i2 = 16-1
t = i-i s = 16-1 x i-i2 = 19-481
8s = 19-481 — 16-1 = 3-381 U = i-i — i = -i
\ &s 3-381
and (average) v = ^ = ^- = 33-81
i. e., the average velocity over the interval of time i to i-i sec.
is 33-81 ft. per sec. This value must be somewhere near the velocity
at the end of the first second, but it cannot be the absolute value,
since even in the short interval of time, viz., -i sec., the velocity has
been increased by a measurable amount. A better approximation
will evidently be found if the time interval is narrowed to -01 sec.
Then — t — i s = 16-1
t = i-oi s = 16-1 X i-oi2 = 16-42361
Ss = -32361 St = -oi
8s -32361
(average) v = ^ = -^~ = 32-361
A value still nearer to the true will be obtained if the time
interval is made -ooi sec. only.
t = i s = 16-1
t = i-ooi s = 16-1 X i-ooi2 = 16-1322161
8s = -0322161 8^ = -ooi
8s -0322161
and (average) v = ^ = ~^^— = 32-2161
By taking still smaller intervals of time, more and more nearly
correct approximations would be found for the velocity ; the values
of v all tending to 32-2, and thus we are quite justified in saying that
when / = i, v = 32-2 ft. per sec.
Or, using the language of p. 458 (Mathematics for Engineers, Pt. 1),
we state that the limiting value of v as t approaches i is 32-2 ; a
result expressed in the shorter form
(average) v -» 32-2 as 8/ -> o when t = i
where the symbol -> means " approaches "
but (average) v = •£, and thus ~ -> 32-2 as St -> o when t = i
ot ot
INTRODUCTION TO DIFFERENTIATION
Again, an actual velocity is an average velocity over an extremely
small interval of time ; or, in other words, an actual velocity is the
limiting value of an average velocity, so that — .
(actual) v = (average) v
i.e.,
ds
dt
st-^-o
8t
By similar reasoning it could be proved that the actual velocity
at the end of the 2nd second was 64-4 ft. per sec., and at the end
of the 3rd second the velocity was 96-6 ft. per sec.
This example may usefully be continued a step further, by
calculating the values of the acceleration; this being now possible
since the velocities are known.
Tabulating as before —
1
V
Sv
Si
&v
a = sl
I
32-2
—
—
—
—
—
32-2
I
32-2
2
64-4
—
—
—
—
—
32-2
I
32-2
3
96-6
—
—
—
and we note that the average acceleration is constant and is thus
the actual acceleration.
Our results may now be grouped together in one table, in which
some new symbols are introduced, for the following reason. A
velocity is the rate of change of displacement, and is found by
" differentiating space with regard to time," and an acceleration
is the rate of change of velocity, and hence it is a rate of change
of the change of position, and so implies a double differentiation.
Thus whilst -T- is called the first derivative or differential coefficient
dl)
of s with regard to t, -^ is the first derivative of v with regard to t
and the second derivative of s with regard to t.
TU ds , dv dfds\ ,,.,,, , • n
Then v = -, , and « = TT = -T, -57 , this last form being usually
at at at\atj
d2s
written as ^ (spoken as d two s, dt squared) ; and it denotes that
the operation of differentiating has been performed twice upon s.
10
MATHEMATICS FOR ENGINEERS
The complete table of the values of the velocity and the accelera-
tion reads —
s
i
Ss
st
_Ss
--•GO
st
Sv S*s
~ St ~ SP
O
o
'
—
—
16-1
i
16-1
—
—
—
16-1
i
—
—
—
32-2
i
32-2
—
—
48*3
i
48*3
—
—
—
64-4
2
—
—
—
32-2
i
32-2
—
80-5
i
80-5
—
—
—
144-9
3
—
—
_—
—
~~~ '
The next example refers to a similar case, but is treated from
the graphical aspect.
Example i. — Experiments made with the rolling of a ball down an
inclined plane gave the following results —
t (sees.)
o
I
2
3
s (cms.)
o
20
80
180
Draw curves giving the space, velocity and acceleration respectively
at any time during the period o to 3 sees.
By plotting the given values, 5 vertically and t horizontally, the
" space-time " curve or " displacement " curve is obtained; the curve
being a parabola (Fig. i).
Select any two points P and Q on the curve, not too far apart,
and draw the chord PQ, the vertical QN and the horizontal PN.
Then the slope of the chord PQ =
Now PN may be written as St since it represents a small addition
to the value of t at P : also QN = 8s,
8s
so that— slope of chord PQ = ^
but ^ — average velocity between the times OM and OR, hence
Of
the average velocity is measured by the slope of a chord. Now
let Q approach P, then the chord PQ tends more and more to lie
along the tangent at P, and by taking Q extremely close to P the
chord PQ and the tangent at P are practically indistinguishable
INTRODUCTION TO DIFFERENTIATION
ir
the one from the other; whilst in the limit the two lines coincide.
go
Then since the slope of the chord PQ gives the value of ~ , and
ot
the limiting value of 7 is -^ , it follows that the slope of the tangent
ot (it
expresses -^ ; but the slope of a curve at any point is measured
.30
1-5 2-O
Values of hmc
FIG. i.
2-5
by the slope of its tangent at that point, and hence we have evolved
the most important principle, viz., that differentiation is the deter-
mination of the slopes of curves.
[Incidentally it may be remarked that here is a good illustration
of the work on limiting values ; for the slope of a curve, or of the
tangent to the curve, is the limiting value of the slope of the chord,
*'. e., the value found when the extremities of the chord coincide;
and this value does not take the indeterminate form C), as might at
o
first sight be supposed, but is a definite figure.]
12 MATHEMATICS FOR ENGINEERS
Thus the slope of the tangent at any point on the space-time
curve measures the actual rate of change of the space with regard
to time at that particular instant; or, in other words, the actual
velocity at that instant. Hence by drawing tangents to the space-
time curve at various points and calculating the slopes, a set of
values of the velocity is obtained : these values are then plotted
to a base of time and a new curve is drawn, which gives by its
ordinate the value of the velocity at any time and is known as the
" velocity-time " curve.
Since this curve is obtained by the calculation of slopes, or
rates of change, it is designated a derived or slope curve; the
original curve, viz., the space-time curve, being termed the
primitive.
In the case under notice the velocity-time curve is a sloping
straight line, and in consequence its slope is constant, having the
value 40. Hence the derived curve, which is the acceleration-time
curve, is a horizontal line, to which the ordinate is 40. There are
thus the three curves, viz., the primitive or space-time curve, the
first derived curve or the velocity-time curve, and the second derived
curve or the acceleration-time curve.
Graphic Differentiation. — The accurate construction of slope
curves is a most tedious business, for the process already described
necessitates the drawing of a great number of tangents, the calcu-
lations of their respective slopes and the plotting of these values.
There are, however, two modes of graphic differentiation, both of
which give results very nearly correct provided that reasonable
care is taken over their use.
Method i (see Fig. 2). — Divide the base into small elements,
the lengths of the elements not being necessarily alike, but being
so chosen that the parts of the curve joining the tops of the con-
secutive ordinates drawn through the points of section of the base
are, as nearly as possible, straight lines. Thus, when the slope
of the primitive is changing rapidly, the ordinates must be close
together; and when the curve is straight for a good length, the
ordinates may be placed well apart. Choose a pole P, to the left
of some vertical OA, the distance OP being made a round number
of units, according to the horizontal scale. Erect the mid-ordinates
for all the strips.
Through P draw PA parallel to ab, the first portion of the
curve, and draw the horizontal Ac to meet the mid-ordinate of the
first strip in c. Then dc measures, to some scale, the slope of the
chord ab, and therefore the slope of the tangent to the primitive
INTRODUCTION TO DIFFERENTIATION 13
curve at m, or the average slope of the primitive from a to b, with
reasonable accuracy.
Continue the process by drawing PM parallel to bl and Ms
horizontal to meet the mid-ordinate of the second strip in s : then
cs is a portion of the slope or derived curve.
Repeat the operations for ah1 the strips and draw the smooth
curve through the points c, s, etc. : then this curve is the curve
of slopes.
FIG. 2. — Graphic Differentiation, Method i.
Indicate a scale of slope along a convenient vertical axis and
the diagram is complete : the scale of slope being the old vertical scale
divided by the polar distance expressed in terms of the horizontal units.
E. g., if the original vertical scale is i" = 40 ft. Ibs. and the
horizontal scale is i" = 10 ft. : then, if the polar distance p is taken
as 2", i. e., as 20 horizontal units,
the new vertical scale, or scale of slope, is i" = ^ — ' — 2 Ibs.
20 ft.
Proof of the construction. —
The slope of the primitive curve at m = slope of curve ab
= bf = OA. = cd
~af~ p " p
14 MATHEMATICS FOR ENGINEERS
or, the ordinate dc, measured to the old scale, = p x the slope of
the curve at m.
If, then, the original vertical scale is divided by p the ordinate
dc, measured to the new scale, = slope of the curve at m.
The great disadvantage of this method is that parallels have
to be drawn to very small lengths of line and a slight error in the
setting of the set square may quite easily be magnified in the draw-
ing of the parallel. Hence, for accuracy, extreme care in draughts-
manship is necessary.
It should be observed that this method of graphic differentiation
is the converse of the method of graphic integration described in
FIG. 3. — Graphic Differentiation, Method 2.
Chapter VII (Part I), and referred to in greater detail in Chapter V
of the present volume.
Method 2. — Let ABC (Fig. 3) be the primitive curve.
Shift the curve ABC forward to the right a horizontal distance
sufficiently large to give a well-defined difference between the
curves DEF and ABC ; but the horizontal distance, denoted by h,
must not be great. From the straight line base OX set up ordinates
which give the differences between the ordinates of the curves
ABC and DEF, the latter curve being treated as the base : thus
ab = a'b'. Join the tops of the ordinates so obtained to give the
new curve G6H, and shift the curve G6H to the left a horizontal
distance = -, this operation giving the curve MPN, which is the
true slope or derived curve of the primitive ABC. Complete the
diagram by adding a scale of slope, which is the old vertical scale
divided by h (expressed in horizontal units).
INTRODUCTION TO DIFFERENTIATION
This method can be still further simplified by the use of tracing
paper, thus : Place the tracing paper over the diagram and trace
the curve ABC upon it; move the tracing paper very carefully
forward the requisite amount, viz., h, and with the dividers take
the various differences between the curves, such as a'b'. Step off
these differences from OX as base, but along ordinates - units re-
45
moved to the left of those on which the differences were actually
measured : then draw the curve through the points and this is
the slope curve.
Tamp (C*)
JO 4O SO 6O O
h" 20 H "Tung (mins.)
FIG. 30. — Variation of Temperature of Motor Field Coils.
Examples on the use of these two methods now follow.
Example 2. — The temperature of the field coils of a motor was
measured at various times during the passage of a strong current, with
the following results —
Time (mins.) .
o
5
10
15
20
25
3°
35
40
45
50
55
60
65
Temperature (C.°)
20
26
32-5
4i
46
49
52-5
54-5
56-5
58
59-5
61
61-7
62
Draw a curve to represent this variation of temperature, and a
curve to show the rate at which the temperature is rising at any instant
during the period of 65 mins.
The values of the temperature when plotted to a base of time give
the primitive curve in Fig. 30.
i6
MATHEMATICS FOR ENGINEERS
To draw the slope curve we first divide the base in such a way that
the portions of the curve between consecutive ordinates have the same
inclination for the whole of their lengths, i. e., the elements of the curve
are approximately straight lines. Thus, in the figure, there is no
appreciable change of slope between A and a, or a and d. There is
no need to draw the ordinates through the points of section for their
full lengths, since the intersections with the primitive curve are all that
is required. Next a pole P is chosen, 20 horizontal units to the left
of A, and through P the line PB is drawn parallel to the portion of the
curve Aa. A horizontal B6 cuts the mid-ordinate of the first strip
at b, and 6 is a point on the slope or derived curve. The processes
repeated for the second strip, PC being drawn parallel to ad and Cc
drawn horizontal to meet the mid-ordinate of the second strip in c,
which is thus a second point on the slope curve. A smooth curve
through points such as b and c is the slope curve, giving by its ordinates
the rate of increase of the temperature ; and it will be observed that
the rate of increase is diminished until at the end of 65 mins. the rate
of change of temperature is zero, thus indicating that at the end of
65 mins. the losses due to radiation just begin to balance the heating
effect of the current.
Since the polar distance = 20 units, the scale of slope
_ original vertical scale
20
and in the figure the original vertical scale is i" = 20 units ; hence the
scale of slope is i" = i unit; and this scale is indicated to the right of
the diagram.
Example 3. — Plot the curve y = x"-, x ranging from o to 3, and use
Method 2 to obtain the derived curve.
The values for the ordinates of the primitive curve y — x"- are as
in the table —
X
o
I
2
3
y
o
I
4
9
and the plotting of these gives the curve OAB in Fig. 4.
Choosing h as -5 horizontal unit, the curve is first shifted forward
this amount, and the curve CG results. The vertical differences be-
tween these two curves are measured, CG being regarded as the base
curve, and are then set off from the axis of x as the base. Thus when
x — 3, the ordinate of the curve OAB is 9 units, and that of CG is 6-25,
so that the difference is 2-75, and this is the ordinate of the curve MN.
By shifting the curve MN to the left by a distance = — , i. e., -25 hori-
zontal unit, the true slope curve ODE is obtained : this is a straight
INTRODUCTION TO DIFFERENTIATION
17
line, as would be expected since the primitive curve is a " square "
parabola.
As regards the scale of slope, the new vertical scale
_ old vertical scale,
and since h = -5, the new vertical scale, or scale of slope, which is used
when measuring ordinates of the curve ODE, is twice the original
vertical scale.
The derived curve supplies much information about the primi-
tive. Thus, when the ordinate of the derived curve is zero, i. e.,
when the derived curve touches or cuts the horizontal axis, the
O -5 / 1-5 2. 2-5
FIG. 4. — Graphic Differentiation.
slope of the primitive is zero; but if the slope is zero the curve
must be horizontal, since it neither rises nor falls, and this is the
case at a turning point, either maximum or minimum. Hence turning
points on the primitive curve are at once indicated by zero ordinates
of the slope curve.
Again, a positive ordinate of the derived curve implies a positive
slope of the primitive, and thus indicates that in the neighbourhood
considered the ordinate increases with increase of abscissa. Also
a large ordinate of the slope curve indicates rapid change of ordinate
of the primitive with regard to the abscissa.
This last fact suggests another and a more important one. By
a careful examination of the primitive curve we see what is actually
c
iS
MATHEMATICS FOR ENGINEERS
happening, whilst the slope curve carries us further and tells us
what is likely to happen. In fact, the rate at which a quantity is
changing is very often of far greater importance than the actual
value of the quantity; and as illustrations of this statement the
following examples present the case clearly.
Example 4. — The following table gives the values of the displacement
of a 21 knot battleship and the weight of the offensive and defensive
factors, viz., armament, armour and protection. From these figures
(2000 _
TOGO
2OOOO
84OOO £60OO E8OOO 3OOOO
Values of P
FIG. 5. — Displacement and Armament of Battleship.
calculate values of Q (ratio of armament, etc., to displacement) and q
(rate of increase of armament, etc., with regard to displacement).
Find also the values of ..
Displacement \
P tons J
18000
2000O
22OOO
24000
2600O
28000
30000
Armament, etc., \
p tons /
6880
7850
8830
9820
10810
11820
12845
INTRODUCTION TO DIFFERENTIATION
The values of Q are found by direct division and are —
p
18000
20000
22OOO
24000
26OOO
28OOO
3OOOO
Q
•383
•392
•40!
•409
•416
•422
•428
dp
Values of q, i. e., -,£, may be found by (a) construction of a slope
curve, or (6) tabulating differences.
(a) By construction of a slope curve. — Plotting p along the vertical
axis and P along the horizontal axis (see Fig. 5), we find that the points
lie very nearly on a straight line. Hence the slope curve is a hori-
zontal line, whose ordinate everywhere is the slope of the original
line. By actual measurement the slope is found to be -498 and thus
•498. This
di)
the average value for ~~, over the range considered, is
average value of the rate of change does not, however, give as much
information for our immediate purpose as the separate rates of change
considered over the various small increases in the displacement.
(6) By tabulation of differences, as in previous examples —
p
P
sp
SP
, Sp
? SP
18000
6880
_
_
_
—
—
97°
2OOO
•485
2OOOO
7850
—
—
—
—
—
980
2OOO
.490
22OOO
8830
—
—
—
—
990
2OOO
•495
24000
9820
— '. —
—
—
990
2OOO
•495
26OOO
I08IO
—
—
—
—
1010
2OOO .505
28OOO
II820
—
—
—
—
1025
2OOO
•5125
3OOOO
12845
—
—
—
JV *
Now sfs = rate of increase of armament compared with displace-
oJr
ment ; as the displacement increases it is seen from the table of values
that this ratio increases, and the questions then arise : " Does this
increase coincide with an increase or a decrease in the values of Q,
and if with one of these, what is the relation between the two changes ? "
By tabulating the corresponding values of q and Q and calculating
the values of i, we obtain the following table (the values of Q at 19000,
21000, etc., being found from a separate plotting not shown here) —
20
MATHEMATICS FOR ENGINEERS
p
19000
21000
23000
25000
27000
29OOO
p
•485
•490
•495
•495
•505
•5125
Q
•387
•396
•405
.412
•419
•426
q
g
1-253
1-236
1-223
1-202
1-203
I-2O4
It will be seen by examination of this table that the fraction Q
decreases as ships are made larger : in other words, while the arma-
ment increases with the displacement, the increase is not so great as
it should be for the size of the ship, since the weight of the necessary
engines, etc., is greater in proportion to the weight of armament and
protection for the larger than for the smaller ships.
Thus, other things being equal, beyond a certain point it is better
to rely on a greater number of smaller ships than a few very large
ones.
Example 5. — Friend gives the following figures as the results of
tests on iron plates exposed to the action of air and water. The
original plates weighed about 2-5 to 3 grms.
Plot these figures and obtain the rate curves for the two cases,
these curves being a measure of the corrosion : comment on the results.
Time in days . ' .
2
7
13
19
26
32
37
In the light : loss of \
weight in grms. . J
•0048
•031
•0645
•08
•093
•126
—
In the dark : loss of)
weight in grms. . /
•0032
•0208
•037
•058
•0674
•0816
•0916
The two sets of values are plotted in Fig. 6, the respective curves
being LLL for the plates exposed in the light, and DDD for those
left in the dark. The effect of the action of light is very apparent
from an examination of these curves. Next, the slope curves for the
two cases are drawn, Method 2 being employed, but the intermediate
steps are not shown. The curve /// is the slope curve for LLL, and
ddd that for the curve DDD.
It will be observed that in both cases the rate of loss is great at
the commencement, but decreases to a minimum value after 20 days
exposure in the case of t the curve ///, and after 25 days in the case of
the curve ddd.
After these turning points have been reached the rate quickens,
INTRODUCTION TO DIFFERENTIATION
21
the effect being very marked for the plates exposed to the light ; and
for these conditions the slope curve /// suggests that the corrosive
action is a very serious matter, since it appears that the rate of loss
must steadily increase.
A further extremely good illustration of the value of slope curves
is found in connection with the cooling curves of metals. In the
early days of the research in this branch of science, the cooling
curve alone was plotted, viz., temperatures plotted to a base of
o s 10 15 £O 25 30
FIG. 6. — Tests on Corrosion of Iron Plate.
time. Later investigations, however, have shown that three other
curves are necessary, viz., an inverse rate curve, a difference curve
and a derived differential curve ; the co-ordinates for the respective
curves being —
(a) Temperature (0) — time (f) curve; t horizontal and 0 vertical.
(b) Inverse rate curve : -, horizontal and 0 vertical. To obtain
this curve from curve (a), the slopes must be very carefully calcu-
lated, and it must be remembered that these slopes are the measures
of the inclinations to the vertical axis and not to the horizontal,
dt ^ft
i. e., are values of -^ and not
,,.
at
22
MATHEMATICS FOR ENGINEERS
(c) 0 vertical and 0— 0X horizontal : 0—0! being the difference
of temperature between the sample and a neutral body cooling under
identical conditions.
(d) 0 vertical and JA horizontal : this curve thus being
the inverse rate curve of curve (c).
Exercises 1. — On Rates of Change and Derived Curves
1. What do the fractions ^- and -,- actually represent (s being a
rV (f(
displacement, and t a time) ? Take some figures to illustrate your
answer.
2. Further explain the meanings of -~- and -r by reference to a
graph.
3. When an armature revolves in a magnetic field the E.M.F.
produced depends on the rate at which the lines of force are being
cut. Express this statement in a very brief form.
4. For a non-steady electric current the voltage V is equal to the
resistance R multiplied by the current C plus the self-inductance L
multiplied by the time rate at which the current is changing. Express
this in the form of an equation.
5. At a certain instant a body is 45-3 cms. distant from a fixed
point. 2-14 seconds afterwards it is 21-7 cms. from this point. Find
the average velocity during this movement. At what instant would
your result probably measure the actual velocity ?
6. At 3 ft. from one end of a beam the bending moment is 5 tons ft.
At 3' 2\" from the same end it is 5-07 tons ft. If the shear is measured
by the rate of change of bending moment, what is the average shearing
force in this neighbourhood ?
7. Tabulate the values of q, i. e., ^ for the following case, the figures
referring to a battleship of 23 knots.
p
18000
2OOOO
22OOO
24OOO
26OOO
28OOO
3OOOO
32OOO
p
6170
7080
8OOO
8930
9890
10855
II820
I28lO
8. Tabulate the values of ^r for the case of a battleship of 25 knots
from the following —
P
18000
2OOOO
22OOO
24000
26OOO
28OOO
3OOOO
32OOO
p
5210
6050
6910
7790
8660
9550
10460
II370
= sp and Q = F
INTRODUCTION TO DIFFERENTIATION
9. Tabulate the values of the velocity and the acceleration for
the following case —
Space (feet)
i
2-4
4'4
6
7-6
II-2
15-6
2O'4
Time (sees.)
•2
•4
•6
•7
•8
I
1-2
i-4
10. Plot the space-time curve for the figures given in Question 9
and by graphic differentiation obtain the velocity -time and the accelera-
tion-time curves.
11. Plot the curve y — -$x3 from x = —2 to x = +4 and also its
derived curve. What is the ordinate of the latter when x — 1-94?
12. Given the following figures for the mean temperatures of the
year (the average for 50 years), draw a curve for the rate of change of
temperature and determine at what seasons of the year it is most
rapid in either direction.
Time (intervals of J month)
o
i
2
3
4
5
6
7
8 9
10
ii
Temperature
38-6
37'9
38-4
39-8
38-5
39-5
4°-3
40-7
4I-5
45-5
45-5
48-5
12
49-3
13
14
15
16
I?
IS
19
20
21
22
23
24
25
26
52
55
57-2
58-4
60-5
61-4
62-5
62-9
62-2
62-5
61-1
59-8
58-2
55-8
27
54-2
28
51
29
48-8
30
46-8
3*
43-5
32
42-1
33
40-6
34
39-8
35
38-8
36
38-6
13. s is the displacement from a fixed point of a tramcar, in time
/ sees. Draw the space-time, velocity-time and acceleration-time curves.
t
0
.1 2
3
4
5
6
7
8
9
10
s
o
4 ii
21
34
50
69
9i
116
144
175
The scales must be clearly indicated.
14. The table gives the temperature of a body at time / sees, after
it has been left to cool. Plot the given values and thence by differ-
entiation obtain the rate of cooling curve. What conclusions do you
draw from your final curve ?
Time (mins.)
o
I
2
3
4
5
6
7
8
9
Temp. (F.°)
136
134
132
130
128
126-5
124-8
123-3
122
120-5
—
10
ii
12
13
14
15
16
17
18
19
119-3
118
II6-8
II5-5
114-5
II3-5
112-5
111-5
110-5
109-5
MATHEMATICS FOR ENGINEERS
15. The following figures give the bending moment at various
points along a beam supported at both ends and loaded uniformly.
Draw the bending moment curve, and by graphic differentiation obtain
the shear and load curves. Indicate clearly the scales and write down
the value of the load per foot run.
Distance from one \
end (ft.) . . /
0
2
4
6
8
10
12
14
16
18
20
Bending moment \
(tons ft.) . . J
o
3'5
6-3
8-4
9-6
10
9-6
8-4
6-3
3'5
O
16. By taking values of 0 in the neighbourhood of 15° find the
actual rate of change of sin 6 with regard to 0 (0 being expressed in
radians). Compare your result with the value of cos 15°. In what
general way could the result be expressed ?
17. If the shear at various points in the length of a beam is as in
the table, draw the load curve (i. e., the derived curve) and write down
the loading at "3^ ft. from the left-hand end.
Distance from left-hand end (ft.)
0
I
2
3
4
5
6
Shearing force (tons) ....
0
•i
•3
•6
i
i'5
2-1
18. An E.M.F. wave is given by the equation
E = 150 sin 314^ + 50 sin 942^.
Derive graphically the wave form of the current which the E.M.F. will
send through a condenser of 20 microfarads capacity, assuming the
condenser loss to be negligible.
dE
Given that C = K~j7' where C is current and K is capacity.
19. If momentum is given by the product of mass into velocity,
and force is defined as the time rate of change of momentum, show
that force is expressed by the product of mass into acceleration.
20. The following are the approximate speeds of a locomotive on
a run over a not very level road. Plot these figures and thence obtain
a curve showing the acceleration at any time during the run.
Time (in mins. and sees.)
o
i-o
2-15
6-15
9-22
n-45
14-26
16-33
20-52
23-10
Speed (miles per hour) .
start
6
10
18-2
22-8
25-5
28
29-2
28-6
26-1
21. Taking the following figures referring to CO2 for use in a re-
frigerating machine, draw the rate curve and find the value of -£- when
ctt
t=i8° F.
t° F. . '
5
o
5
10
15
20
25
30
35
40
£>lbs.per\
sq. in J
285
310
335
363
392
423
456
491
528
567
INTRODUCTION TO DIFFERENTIATION
22. The weight of a sample of cast iron was measured after various
heatings with the following results; the gain in weight being due to
the external gases in the muffle.
Number of heats . .
o
2
6
12
22
23
24
25
26
Weight ....
146-88
146-94
147-04
H7-54
148-02
I48-II
I48.27
148-36
148-46
-
27
30
35
39
45
148-61
149-18
150-49
152-36
156-44
Plot a curve to represent this table of values, and from it construct
the rate curve.
23. The figures in the table are the readings of the temperature of
a sample of steel at various times during its cooling. Plot these values
to a time base, and thence draw the " inverse rate " curve, i. e., the
curve in which values of -«. are plotted horizontally and the temperatures
along the vertical axis.
Time in sees. (I) . .
75 j 9°
i°5
120
135
150
165
180
195
210
225
Temperature in C.c (0)
1
850 848
844-7
842
839-5
838-5
838-2
838-1
838
837-9
837-5
240
255
270
285 292-5
300
315
330
345
360
367-5
375
350
4°5
836
833
829
825 8-'3-3
822-2
821-7
821-5
821-3
821-1
819
8i5
813
8II-6
CHAPTER II
DIFFERENTIATION OF FUNCTIONS
Differentiation of ax". — It has been shown in Chapter I
how to compare the changes in two quantities with one another,
and thus to determine the rate at which one is changing with
regard to the other at any particular instant, for cases in which
sets of values of the two variables have been stated. In a great
number of instances, however, the two quantities are connected
by an equation, indicating that the one depends upon the other,
or, in other words, one is a function of the other. Thus if y = 5#3,
y has a definite value for each value given to %, and this fact is
expressed in the shorter form y =f(x). Again, if z — ijx^y—^xy3
+5 log y> where both x and y vary, z depends for its values on
those given to both x and y, and z — f(x, y).
To differentiate a function it is not necessary to calculate
values of x and y and then to treat them as was done to the given
sets of values in the previous chapter. This would occasion a
great waste of time and would not give absolutely accurate results.
Rules can be developed entirely from first principles which permit
the differentiation of functions without any recourse to tables of
values or to a graph.
We now proceed to develop the first of the rules for the differen-
tiation of functions ; and we shall approach the general case, viz.,
that of y — axn, by first considering the simple case of y = x3.
Our problem is thus to find the rate at which y changes with regard
to x, the two variables (y the dependent and x the independent
variable or I.V.) being connected by the equation y = xs.
The rate of change of y with regard to x is given by the value
dv
of -j-, and this is sometimes written as Dy when it is clearly
understood that differentiation is with regard to x : the operator D
having many important properties, as will be seen later in the
book. If y is expressed as f(x), then -p is often written JfQ
fix dx
or /'(*).
26
DIFFERENTIATION OF FUNCTIONS 27
~, Dy, -^-* or f'(x) is called the derivative or differential
coefficient of y with respect to x; and the full significance of the
latter of these terms is shown in Chapter III.
We wish to find a rule giving the actual rate of change of y
with regard to x, y being = x3, the rule to be true for all values
of x. As in the earlier work, the actual rate of change must be
determined as the limiting value of the average rate of change.
Let x be altered by an amount Sx so that the new value of
x = x -f- 8x ; then y, which depends upon x, must change to a
new value y + Sy, and since the relation between y and x is y — x3
for all values of x —
(new value of y) = (new value of x)3
or y+Sy = (x+Sx)3 = ,r3+3#2 . 8x+$x . (Sx)2+($x)3 (i)
but y = x3 ........... (2)
Hence, by subtraction of (2) from (i) —
y+Sy-y = 3*2 . Sx+3x(Sx)2+(Sx)3
and Sy = $x2 . Sx+3x(Sx)2+(Sx)3.
Divide through by Sx, and —
Thus an average value for the rate of change over a small
interval Sx has been found; and to deduce the actual rate of
change the interval Sx must be reduced indefinitely.
Let Sx=-ooi; then ^ = 3*2+ (3* X-ooi) + -000001
-oooooi
whilst if 8x = -ooooi—
jh_j
:- = 3#2-f -00003^;+ -oooooooooi . (3)
Evidently, by still further reducing &x the 2nd and 3rd terms
of (3) can be made practically negligible in comparison with the
ist term.
Then, in the limit, the right-hand side becomes 3*2,
and thus- ^ = T f = 3*2
dx i ^Sx
dx- = *X
28
MATHEMATICS FOR ENGINEERS
This relation can be interpreted graphically in the following
manner : If the curve y = xs be plotted, and if also its slope curve
be drawn by either of the methods of Chapter I, then the equation
to the latter curve is found to be y = $x2.
The two curves are plotted in Fig. 7.
10
01234
X
FIG. 7. — Primitive and Slope Curves.
Example i. — Find the slope of the curve y — x3 when x = 4.
dy dx3 o
The slope of the curve = -v1 = -j— = 3*
CLX CLX
and if x = 4 -,- = 3 X 42 = 48.
Meaning that, in the neighbourhood of x — 4, the ordinate of the
curve y = x3 is changing 48 times as fast as the abscissa ; this fact
being illustrated by Fig. 7.
Working along the same lines, it would be found that -jj— =
dx5
-y-
himself).
and -- = 5#4 (the reader is advised to test these results for
DIFFERENTIATION OF FUNCTIONS 29
Re-stating these relations in a modified form —
dx3
- = 4^ = 4*
ft*
We note that in all these cases the results take the form —
dxn
— - = nxn ~ \
dx
Thus the three cases considered suggest a general rule, but
it would be unwise to accept this as the true rule without the
more rigid proof, which can now be given.
Proof of the rule —
dxn
— -M Vtt ~~ *
- /(vV •
ax
Let y = xn, this relation being true for all values of x . (i)
If x is increased to x-}-8x, y takes a new value y-f-Sy, and
from (i) it is seen that —
y-|-Sy = (x+8x)n.
Expand (x-\-8x)n by the Binomial Theorem (see p. 463, Part I).
Then —
Subtract (i) from (2), and —
Divide by 8^ —
8v , , n(n—i) „ 0/c. , , «(w — i)(n — 2) M ,/» \9 ,
-•-=«^n-i4- v --- 'xn-2(8x)-\ — s — — 'xn-3(8x)2-t- terms
8,r |_2 |_3
containing products of (Sx}3 and higher powers of (8x).
Let 8* be continually decreased, and then, since Sx is a factor
of the second and all succeeding terms, the values of these terms
can be made as small as we please by sufficiently diminishing Sx.
30 MATHEMATICS FOR ENGINEERS
Thus in the limit — f- -> n xn - *
Sx
dv
or --
Id
Hence the first rule for differentiation of functions is established,
viz. —
i. e., differentiation lowers the power of the I.V. by one, but the
new power of x must be multiplied by the original exponent.
The reason for the multiplication by the n can be readily seen,
for the bigger the value of n the steeper is the primitive curve
and therefore the greater the change of y for unit change of x.
The n actually determines the slope of the primitive (cf. Part I,
p. 340), and it must therefore be an important factor in the
result of differentiation, since that operation gives the equation of
the slope curve.
To make the rule perfectly general, aUowance must be made
for the presence of the constant multiplier a in axn.
It will be agreed that if the curve y = x3 had been plotted,
the curve y = ^x3 would be the same curve modified by simply
multiplying the vertical scale by 4. Hence, in the measurement
of the slope, the vertical increases would be four times as great
for the curve y = 4#3 as for the curve y = x3, provided that the
same horizontal increments were considered.
Now the slope of the curve y = x3 is given by the equation —
dy - ix*
dx~3X
so that the slope of the curve y = ^x3 is given by —
^ = 4x3*2 = I2x2.
In other words, the constant multiplier 4 remains a multiplier
throughout differentiation. This being true for any constant
factor —
j-axa = nax
H-l
dx*
Accordingly, a constant factor before differentiation remains as
such after differentiation.
We can approach the differentiation of a multinomial expression
DIFFERENTIATION OF FUNCTIONS
by discussing the simple case y = $x2 + 17 (a binomial, or two-
term expression). The curves y = $x2 and y = $x2 + 17 are seen
plotted in Fig. 8, and an examination shows that the latter curve
is the former moved vertically an amount equal to 17 vertical
units, i. e., the two curves have the same form or shape and
consequently their slopes at corresponding points are alike. Thus
if a tangent is drawn to each curve at the point for which x = 2-5,
the slope of each tangent is measured as — , i.e., 25; and con-
sequently the diagram informs us that the term 17 makes no
difference to the slope.
JOO.
FIG. 8.
Thus —
dx-
Now, by differentiating 5#2
dx^
17 term by term, we have —
since 17 is a constant and does not in any way depend upon x,
and therefore its rate of change must be zero.
It is seen that in this simple example it is a perfectly logical
procedure to differentiate term by term and then add the results;
and the method could be equally well applied to all many-term
expressions.
32 MATHEMATICS FOR ENGINEERS
Hence— jx(axn+bxn~'i+cxn-2+ . . . d)
= naxn~l+b(n — i)xn~2+c(n — 2)xn~3 + . . .
and ^x(axa-{-b) = nax"-1.
To apply these rules to various numerical examples : —
Example 2. — Differentiate with respect to x the function —
= (gx i-6;r6)
= i4'4#-8+#^'5 or IA-AX-*-\ — — _
V x
Example 3.— If y = -Sx AA, find the value of ^.
' *
y = -8x A/L = -8x~, = -8^~^ or
V ^s ^
so that in comparison with the standard form —
a = -8 and n = —1-5.
Then—
—
or _
Example 4. — If /w1'41 = C, the equation representing the adiabatic
expansion of air, find -—-.
In this example we have to differentiate p with regard to v, and
before this can be done p must be expressed in terms of v.
Now pv1-*1 = C, so that p = ^ = Cw-1'11.
Hence ft = ^-Cz;-1-11 = Cx -i-4iy-2-11 = - i-4iCw-2-41
dv dv
and this result can be put into terms of p and v only, if for C we write
its value pvl'tl.
Thus = -i-4ix/>w1-4lx»-8'll= —i-4ipv-1= - -'.
DIFFERENTIATION OF FUNCTIONS 33
Example 5. — The formula giving the electrical resistance of a length
of wire at temperature t° C. is —
where R0 = Resistance at o° C. Find the increase of resistance per
i° C. rise of temperature per ohm of initial resistance, and hence state
a meaning for «.
The question may be approached from two standpoints ; viz. —
(a) Working from first principles.
"D _ "O "D •[
i. e., increase in resistance for t° C. =
but this is the resistance increase for initial resistance R0, hence
T? a
increase in resistance per i° C. per ohm initial resistance = -^- = a.
(&) By differentiation.
Rate of change of R« with regard to t = -~
0a = R0a
and consequently the rate of change of resistance per i° C. per i ohm
initial resistance = o.
The symbol a is thus the " temperature coefficient," its numerical
value for pure metals being -0038.
Example 6. — Find the value of -^-(45*— \+6sl — 1-84).
CIS \ S '
Write the expression as 45*— 3s~2+6s-5— 1-8*.
Then—
2f ' n-l\
Example 7. — If x = an\i — a n /, a formula referring to the flow
dx
of a gas through an orifice, find an expression for -5-.
*( n^\
As it stands an\l — a n ) is a product of functions of the I.V.
(in this case a), and it cannot therefore be differentiated with our
D
34 MATHEMATICS FOR ENGINEERS
present knowledge. We may simplify, however, by removing the
brackets, and then —
2 n-l 2 2 M+l
x — an — aT^ n — an — a n
+ l
, , / 2 n
dx d I •
— = —\an — a »
da da\
_i -i
2 n n-\-I n
= - X a --- a
n • n
2 n
-a -- — a
n n
2-n
7 n .
Example 8. — Determine the value of —
5— -45m9-86
rfm\ 5W
To avoid the quotient of functions of m, divide each term by 5m4'32,
T>7*w»75 .^C*M.^»36
then the expression =
and - (expression) = (3'4X — 3'57^~4'57) — -
~4'" — -499m4-54— 9'
Proof of the construction for the slope curve given on p. 14.
Let us deal first with the particular case in which the equation
of the primitive curve is y = x2.
Referring to Fig. 4, the equation of the curve OAB is y = x2,
and the equation of the curve CG is yl — (x — h)2 = x2 -f h2 — 2xh.
Hence the difference between the ordinates of the curves OAB
and CG, the latter being regarded as the base curve —
= 2xh—hz
so that the equation of the curve MN is
y-5 = 2xh—h2.
DIFFERENTIATION OF FUNCTIONS 35
Now the curve ODE is the curve MN shifted back a distance of
- horizontal units, and hence its equation is y3 = z(x-\ — jh — h2,
2 \ ^/
since for x we must now write f x-\—\
Thus the equation of curve ODE is —
y3 = 2xh
y»
or ~ = 2x
h
A/
i. e., if Y be written for ~, Y = 2x
or the equation of the curve ODE is that of the slope curve of
the curve y ~ x2 provided that the ordinates are read to a certain
scale; this scale being the original vertical scale divided by h
expressed in horizontal units.
Hence the curve ODE is the slope curve of the curve OAB.
Before discussing the general case, let us take the case of the
primitive with equation y = x3.
If the curve be shifted forward an amount = h, the equation
of the new curve is —
yi = (x-W
and the equation of the curve giving the differences of the
ordinates is — •
y2 = y— y^-x3— (x— h)3 = x3—
By shifting this curve - units to the left we change its equation,
by writing (x-\ — J in 'place of x, into the form —
Dividing by h —
h 4
or Y = 3*2+-
36 MATHEMATICS FOR ENGINEERS
hz
Now if h is taken sufficiently small, — is negligible in comparison
with 3#2, and we thus have the equation of the curve Y = 3#2,
which is the slope curve of the curve y = x3 ; but the ordinates
must be measured to the old vertical scale divided by h.
We may now consider the case of the primitive y = xn. Adopt-
ing the notation of the previous illustrations —
n(*L-ll)xn-2}l2_ . _ \
\
Write (*+2) in place of x, and then
h L 2 8
n(n— i) „ 27 «(w— i)(w— 2) n_3,2
_ ._, _> _ —^C^1 — ft _ - _ OC — ft —~
2 4
= nXn-1-}- terms containing A as a factor.
Hence if h is made very small —
Y or = w*"-1.
Exercises 2. — On Differentiation of Powers of the I.V.
1. Find from first principles the differential coefficient of x*.
o
2. Find the slope of the curve y = —2 when x = -5
•^
(a) By actual measurement and (6) by differentiation.
3. The sensitiveness of a governor is measured by the change of
height corresponding to the change of speed expressed as a fraction
of the speed. Thus if h and v represent respectively the height and
dl)
speed, the sensitiveness — dh -. --- . If the height is inversely pro-
portional to the square of the velocity, find an expression for the
sensitiveness.
Differentiate with respect to x the functions in Exs. 4 to 15.
- --*» ™ '2I5
4. 3*9. 5. -. 6. 8i-5*-*». 7. igx™. 8.
A/-O »/ " - ^" ' .£ ,J
DIFFERENTIATION OF FUNCTIONS
37
9.
8*'
10.
«•
12.
(*3-7)2-8
~
13.
14.
15.
16. Find the value of -]- when pv1-3 = 570 and v = 28-1.
.
17. Find the value of
Jv
18. If E = — I5+I4T — -oo68T2, find the rate of change of E with
regard to T when T has the value 240.
, dH , dtt i f dp , \
19. Calculate the value of -j- from -T- = -- \v^-+yp( when
dv <fo 7— 1 1 dv -^)
pv1-3 = C and y = 1-4.
20. Find the rate of discharge ( -,- J of air through an orifice from a
tank (the pressure being 55 Ibs./n") from the following data —
I44/>V = wRT
R = 53-2, V = 47-7, T = 548.
Time (sees ) (/)
o
60
I35
21$
3IS
Pressure (Ibs. per sq. in.) (p)
63
45
30
15
10
Hint. — Plot p against / and find ? when p = 55.
21. If P = load displacement of a ship,
p = weight of offensive and defensive factors.
Then P = aP+bP*+p.
Find the rate of increase of armament and protection in relation
to increase of displacement.
II \ O / \ I \ *>
yai [ I __ y \ *u\t I 4f \ 901 1 \) V I *
22. if M = w(t xw(I+y)-w(y *> ,
2/ \ // 2
constants.
\Vv ('/'M
23. If M = — ^(/2 — 4y2), find the value of y that makes ,— = o.
2/2 v dy
n* Tt c* w((x-}-ny)2 x2} ,. , , , , dS
24. If S = -i s — ST-" k find the value of -,-.
2 I / y J ax
25. Find the value of h which makes -jr- = o when —
dh
dM
""' ^ and
[h is the height of a Warren girder; and the value found will be
the height for maximum stiffness.]
26. If p = -^ — A and q — -75+ A, find the value of v f in terms of
r3 * r3 dr
p and q. (This question refers to the stresses in a thick spherical
shell, p being the radial pressure, and q the hoop tension.)
38 MATHEMATICS FOR ENGINEERS
27. In a certain vapour the relation between the absolute tem-
perature T and the absolute pressure p is given by the equation
T = 140^1 + 465, and the latent heat L is given by L = 1431 — -ST.
Find the volume, in cu. ft., of i Ib. of the vapour when at a pressure
of 81 Ibs. per sq. in. absolute, from —
V .n-? — J • (T — TjRl
I44T dp
28. For a rolling uniform load of length r on a beam of length /,
the bending moment M at a point is given by —
«)«"!// v\ ieiv%
M =
If y is a constant, find an expression for the shear (i. e., the rate
of change of bending moment).
29. Given that p = electrical resistance in microhms per cu. cm.
and x = percentage of aluminium in the steel,
then p = 12 + 12#— -3#2 for steel with low carbon content.
Find the 'rate of increase of p with increase of aluminium when
x = 4.
30. The equation giving the form taken by a trolley wire is —
y =
and the radius of curvature =
2000 1760
i
dx*
Find the value of the radius of curvature.
Good examples of the great advantage obtained by utilising
the rules of differentiation already proved are furnished by the
two following examples, which have reference to loaded beams.
Example g. — Prove that the shearing force at any point in a beam
is given by the rate of change of the bending moment at that point.
Consider two sections of the beam 8x apart (see Fig. 9). The
shear at a section being denned as the sum of all the force to the right
of that section, let the shear at b = S, and let the shear at a = S+8S.
Also let the moment of all the forces to the right of b (i. e., the bending
moment at b) = M, and let the bending moment at a = M+SM.
Taking moments about C —
M+SM = M+(S+8S)**+S(8*)
or 8M =
DIFFERENTIATION OF FUNCTIONS 39
SM , SS
Dividing by Sx, ^- — S-\
and when 8x is diminished indefinitely, SS becomes negligible
and — = S.
S+SS
.^ooooRftooo
s H ""*
FIG. 9. FIG. 10.
Examples on Loaded Beams.
The last example should be considered in conjunction with the
following : —
Example 10. — For a beam of length /, fixed at one end and loaded
uniformly with w tons per foot run, the deflection y at distance x
from the fixed end is given by the formula —
E being the Young's Modulus of the material of the beam, and I being
the moment of inertia of the beam section.
, dy d*y d?y , d*y
Find the values of -~, -j^, , , and -,—..
dx? dxv dx3 dx*
y =
Differentiating, = {(W X 2x) - (4l x 3^2) +4*3}
Differentiating again, g -
40 MATHEMATICS FOR ENGINEERS
Differentiating again, * -
w
d*y _ d(d?y\ _ dJTw
~ ~ ~ (* '
Carrying the differentiation one stage further —
d*y _ d(d?y\
5** ~ ~dx\dx*J
Physical meanings may now be found for these various deriva-
tives. Referring to Fig. 10, consider a section of the beam distant
x from the fixed end. To the right of this section there is a length
of beam l—x loaded with w tons per foot, so that the total load
or total downward force on this length is w(l — x) ', and since this
load is evenly distributed, it may be all supposed to be concentrated
at distance from the section.
2
Now the bending moment at the section
= moment of all the force to the right of the section
/l—x\ w
= force X distance = w(l—x)x( ) = ~(l—x)\
d2v
Comparing this result with the value found for ~z> we notice
that the two are alike except for the presence of the constants
E and I : thus -~ must be a measure of the bending moment.
Actually the rule connecting M, the bending moment, and
its-
dx* 1S
M -tPy „
= i °r M =
the proof of this rule being given in a later chapter.
Again, we have proved in the previous example that the shear
is given by the rate of change of bending moment : thus —
dx dx dx2 dx3
= w(x— /
DIFFERENTIATION OF FUNCTIONS 41
a result agreeing with our statement that the shear at a section is
the sum of all the loads to the right of the section. [The reason
for the minus sign, viz., (x— I), being written where (l—x) might
be expected need not be discussed at this stage.]
Continuing the investigation —
d* w
or v3 = w
dtf
but w is the loading on the beam and —
-~d*y d f^d3y\ dS
EFr4 = j-( El j-4 — j-
dx* dx\ dx3/ dx
so that the loading is measured by the rate of change of the shear.
If now the deflected form is set out, by constructing successive
slope curves we obtain, respectively, the slope curve of the deflected
form, the bending moment curve, the shear curve and finally the
curve of loads.
Example n. — The work done in the expansion of gas in gas turbines
is given by —
where r is the ratio of expansion.
Compare governing by expansion control with governing by
alteration of the initial temperature, from the point of view of
efficiency.
Deal first with the expansion control, i. e., regard Tj as constant
and r as variable. Then the rate at which the work is increased with
respect to r is -3—.
Now
3—
dW
w
42 MATHEMATICS FOR ENGINEERS
Now regard r as constant, but T, as variable.
/7W *? P V / w-l\
Then- ^ = ^^rV-^)
u J-! n — i J-o
and, expressing the two results in the form of a ratio —
/7W /fW PVT (w r^T
«vv . «vv _ x-1v0±1 {n i)J-0
-( ""^
^yn\I— r n I
Lengths of Sub-tangents and Sub-normals of Curves. —
The projection of the tangent to a curve on to the axis of x is
known as the sub-tangent, i.e., the distance "sub" or "under'?
the tangent. The projection of the normal on the x axis is called
the sub-normal.
The slope of a curve at any point, measured by the slope of
its tangent at that point, is given by the value of -f- there, or if
a = inclination of the tangent to the x axis —
dy
tan a =
dx
In Fig. ii —
FIG. n. — Sub-tangent and Sub- normal.
PA dy
-x-~i = tan a = -f-
AT dx
AT = PA^
ay
But — AT = sub-tangent and PA = y
dx
and hence the length of the sub-tangent = y,-
DIFFERENTIATION OF FUNCTIONS
Again — L APN = a, since L TPN = a right angle
AN sub-normal
tan
i. e.t
or
tan a =
sub-normal
sub-normal = y X tan a = y-~
To find the length of the tangent PT —
(PT)2 = (PA)2+(AT)2
and
In like manner — PN =
FIG. 12.
43
Example 12. — Find the lengths of the sub-tangent and the sub-
normal of the parabola y2 = <\ax (Fig. 12).
yz = AfO-x and y = 2 Va . #*
then
or
Then length of sub-tangent = y
dx
dy
Va
Va
44 MATHEMATICS FOR ENGINEERS
This result illustrates an important property of the parabola and
one useful in the drawing of tangents. For AT = 2.x — 2 X AO, and
hence to draw the tangent at any point P, drop PA perpendicular to
the axis, set off OT = OA and join TP.
The length of the sub-normal AN = y-2-
„, v- Va _ 2 Va Vx Va
— y /\ ___ — _ .
Vx Vx
= 2d.
i. e., the length of the sub-normal is independent of the position of P,
provided that the sub-normal is measured on the axis of the parabola.
Example 13. — Find the lengths of the sub-tangent and the sub-
normal of the parabola — y = i$x2—2x—g
when x = — 2 and also when x — 3.
The axis of this parabola is vertical, and consequently the sub-
normal, which is measured along the x axis when given by the value
of y-^, is not constant.
7dx
Now — y = i5x*—2X—g
dy
and -r- = 30*— 2.
dx
dx 2——
Hence sub-tangent — = y-r- =
-r- —
dy $ox — 2
dv
and sub-normal = y~~ = (i^xz—2x—g) x (30^—2).
ax
Thus when x = — 2
sub-tangent = - ~ — ? _ __g5 umf-s.
_ _ _ —60 — 2 _ 62
sub-normal = (60+4— 9) ( — 62) = —3410 units.
When x = 3
/I35— 6— g\ 120 15
sub-tangent = ( g8 j = -88 = ^ units.
sub-normal = 120 X 88 = 10560 units.
Example 14. — A shaft 24 ft. long between the bearings weighs
2 cwt. per foot run, and supports a flywheel which weighs 3! tons
at a distance of 3 ft. from the right-hand bearing. Find at what
point the maximum bending moment occurs and state the maximum
bending moment.
Regarding the shaft as a simply supported beam AB (see Fig. 13),
we may draw the bending moment diagrams for the respective systems
DIFFERENTIATION OF FUNCTIONS
45
of loading, viz., ADB for the distributed load, being the weight of the
shaft, and ACB for the concentrated load.
The total distributed load is wl, i.e., 24X-I = 2-4 tons, giving
equal reactions of 1-2 tons at A and B; and the bending moment
diagram is a parabola with vertex at D, the maximum ordinate DF
being -5-, i. e., — Q— — or 7-2 tons ft. If for convenience in the
later working the axes of x and y are as shown in the figure, the
equation to this parabola is y2 = <\ax; or taking the value of y as FB
and that of x as DF, 1 22 = 40x7- 2, from which 40 = 20 and
y2 = 20*.
Scale of
Bending Momen
tons- f r
pal lei to AC
FIG. 13.
The load of 3-5 tons produces reactions of — X3*5 tons at B,
24
and — x 3-5 tons at A, i. e., RB = 3-06 and RA = -44 tons : thus the
24
bending moment at E is 3-06x3 = 9-18 tons ft.
Since the total bending moment is obtained by adding the ordinates
of the diagram ADB to the corresponding ordinates of ACB, the
maximum bending moment will be determined when the tangent to
the parabola is parallel to AC, and the position satisfying this condition
can readily be found by differentiation. Thus —
The equation of the curve ADB is y2 = 20* or y = 4-47**, and
the slope of the curve is given by the value of -~.
Now if y = 4-47* ,
= 4-47X^4
e.f
tan « =
_ 4'47
2*3
46
Referring to the figure ACB, tan a = =
and thus- 4^42 = ^i
2** 9'l8
or x* = 1:47X9-18
42
i.e., (DR)» = 4-47X9-18
42
Again, (PR)2 = 2oxDR, and thus PR =
= 4-37 ft.
Thus the maximum bending moment occurs at a distance of
12 — 4-37, i. e., 7-63 ft. from the right-hand bearing.
To find the maximum bending moment —
DR = **/^v*" = -956
\ 42 /
PQ = DF-DR = 7-2--96 = 6-24 tons ft.
Also — ^jLp = — — X9-i8 = 7-16 tons ft.
Hence the maximum bending moment = 7-16+6-24 — 13-4 tons ft.
Exercises 3. — On the Lengths of the Sub-tangent and Sub-normal : also
Beam Problems.
1. Find the lengths of the sub-normal and sub-tangent of the
curve $y = ^x3 at the point for which x = 3.
2. If y = -^W, V = 117, and g = 32-2, find the value of x that
makes the slope of the curve i in 17-4.
3. A parabolic arched rib has a span of 50 ft. and a rise of 8 ft.
Find the equation of the tangent of the slope of the rib. What is the
slope of the tangent at the end ?
4. Find the equation of the tangent to the curve p = ^- at the
v
point for which v — 5.
In Exercises 5 to 7, y is a deflection and x a distance along the
beam. Find, in each case, expressions for the Bending Moment,
Shearing Force and Load. The beam is of uniform section throughout,
and of span /.
5. The beam is supported at both ends, and loaded with W at the
centre.
W llxz xa\
y = -^rT( -=- ) {x is the distance from the centre}.
2EI\ 4 6 /
6. The beam is supported at both ends, and loaded continuously
with w per ft. run.
y — ^^^Y^-g j {x is the distance from the centre}.
DIFFERENTIATION OF FUNCTIONS 47
7. A cantilever loaded with W at the free end.
W/7#2 xa\
y = -p_( --- -j-\ \x is the distance from the fixed end}.
8. Find the lengths of the projections on the y axis of the tangent
and the normal of the parabola, x2 — iob2y + ^c, x having the value ga.
9. Prove that the sub-normal (along the axis of the parabola) of
the parabola x2 — 6y is constant and find the value of this constant.
^ A ™ rr , /- , r> f. A 4.-
10. If El-/ = ---- ,. --- \-C and C = ------- , find the value
rf* 4 6 a 2 12
Differentiation of Exponential Functions. — The rule for
differentiation already given applies only to functions involving
the I.V. (usually the x) raised to some power. A method must
now be found for the differentiation of exponential functions, viz.,
those in which the I.V. appears as exponent ; such as e5x or 4*.
When concerned with the plotting of the curve y = e* (see
Part I, p. 352) mention was made of the fact that, if tangents
are drawn to the curve at various points, the slopes of these
tangents are equal to the ordinates to the primitive curve at the
points at which the tangents touch the curve. Thus the slope
curve of the curve y = ex lies along the primitive, and —
**-- 6*
dx ~
or the rate of change of the function is equal to the value of the
function itself.
We may establish the result algebraically thus —
X % 3C
ex = i-\-x-\ --- [-—
•
1.21.2.31.2.3.4
Assuming that a series composed of an unlimited number of
terms can be differentiated term by term and the results added
to give the true derivative (this being true for all the cases with
which we shall deal), then by differentiation —
dex _
~ f
x3
= ex.
Another respect in which the function e? is unique may be
48 MATHEMATICS FOR ENGINEERS
dx i
noted : the sub-tangerit = y-,- -•- exx— = i, i. e., the sub-tangent
is constant and equal to unity.
The curve y = ex may be usefully employed as a gauge or
template for testing slopes of lines; the curve being drawn on
tracing-paper and moved over the line to be tested until the curve
and line have the same direction, and the ordinate of the curve
being then read, any necessary change of scales being afterwards
made.
The work may now be carried a stage further, so that the rule
for the differentiation of ebx may be found.
Referring to Part I, p. 354, we note that if the curve y = ex
be plotted, then this curve represents also the equation y = ebx if
the numbers marked along the horizontal scale used for the curve
y = ex are divided by b. If, then, the slope of the construction
curve, i. e., that having the equation y = ex, is measured, we can
obtain from it the slope of the curve y — ebx by multiplying the
slope by b, since vertical distances are unaltered, whilst horizontal
distances in the case of y = ebx are T X corresponding horizontal
distances for y = ex.
Hence the slope of the curve y = ebx is b X slope of curve y = ex
r\pbx
or— -p- = bebx
It should be noticed that the power of the function remains
the same after differentiation, but the multiplier of the I.V. becomes
after differentiation a multiplier of the function. This latter rule
must be remembered throughout differentiation, viz., any multiplier
or divisor of the I.V. in the function to be differentiated must become a
multiplier or divisor of the function after differentiation.
From ebx we can proceed to aebx, the result from the differentia-
tion of which is given by —
daebx
dx
= abebx
Example 15. — If y = yrk*t find the value of -f~
dx
-/- = ^~5e~bx = 5 X — T&~^X
dx dxj 6
DIFFERENTIATION OF FUNCTIONS 49
Referring to the last example, note that the power of e is
exactly the same after differentiation as before : the factor — ^
multiplies the I.V. in the original function, and therefore it occurs
as a constant multiplier after differentiation; also the constant
factor 5 remains throughout differentiation.
Example 16. — If C = C0e L, where C and C0 are electrical currents,
R is the resistance of a circuit, L is the self-inductance of the circuit
and t is a time, find the time-rate of change of C.
This example illustrates the importance of the rate of change as
compared with the change itself; for it demonstrates the fact that
for an inductive circuit the change of current is often extremely rapid
and consequently dangerous.
T" * i- t /-> dC, d ~ 5*
Time rate of change of C = -^ = _,XV~ L
dt dt
- C x V?
— ^o Ps T~& L
— -C -~
R^
i. e., the rate of decrease of the current when the impressed E.M.F.
is removed is proportional to the current at the instant the circuit is
broken.
To better illustrate the example, take the case for which the
current at the instant of removal of the E.M.F. is 14-5 amps., the
resistance of the circuit is 6-4 ohms, and its self-inductance -oo6 henry.
Then the rate of change of the current = — -- ^ X 14-5 = — 15470 amps.
*ooo
per second, whereas the actual current is only 14-5 amps.
The expressions ex and ebx are particular forms of the more
general exponential function ax; to differentiate which we may
proceed by either of two methods : —
(a) Working from first principles. In Part I, p. 470, the
expansion for ax is given, viz. —
log a+(*log*)%(*joj^)*
50 MATHEMATICS FOR ENGINEERS
Differentiating term by term —
dax . , . n \2 , /, \sx2
dx = o+log tf+(log a) x+(log a) ~+ . . .
. . (x log a)2 . (x log a)3 ,
= log a{i+x log a+±~- '— > +^— — ' +
L± II.
= log a X ax
da*
(b) Assuming ihe result for the differentiation of ebx —
Let— ax = ebx
so that a — eb, and therefore loge a = b.
d , d
Ta = J-*
ax ax
= logg axax
or ax log a
Then— Tax = Tebx = bebx = loge ax ebx
,, dax
thus — -j— = ax . log a.
d.
Example 17. — Find the value of ^p-
dX
In this case « = 4 and loge 4 = 1-3863.
dAx
Hence -£— = '1-3863 x <\x.
CIX ^^^*^^m**^mm^^^*
Note carefully that this result cannot be simplified by combining
1-3863 with 4 and writing the result as 5-5452*, which is quite incorrect.
The 4 alone is raised to the x power, and 1-3863 is not raised to this
power.
TO
Example 18. — Find the value of , 2(3'^)S-
%
Here a = 3-6 and log 3-6 = 1-2809.
Thus j£(3'6)' = lo§ 3-6 X (3-6)* = 1-2809 x (3-6)*.
Then —
= I -2809 XT- (3-6)* = 1-2809 X 1-2809 x(3-6)j
GfS
= 1-64(3-6)*.
DIFFERENTIATION OF FUNCTIONS 51
Example 19. — Given that s = ^est-\-ye~5tt find the value of -^—255.
s =
= 2O05' — 350 ~5t.
Again — ,.z =
•'• w*~25$ = —
Differentiation of log x. — The rule for the differentiation
of logarithmic functions can be derived either from the expansion
of loge (i + x) into a series, or by assuming the result for the
differentiation of ex. Considering these methods in turn —
(a) Working from first principles. Let y = log x, i. e., loge x.
Then if x be increased to become x-\-8x, y takes a new value
y+Sy, and y+Sy = log (x-{-8x).
(8x\ ( 8x\
i-\ — ) = log x+log ( i-j ),
X / \ X /
therefore —
/ Sx\
(y-j-Sy)— y = log #-|-log ( i-\ — )— log x
\ x /
i. e., 8y = log (l-f*— )
Also log ( i-j — ) can be expanded into a series of the form
\ x /
A- 2 /yO A'1*
log (*+x) = x— -+-—-+ • . . (see Part I, p. 470)
^ o 4
so that —
r ~ I — \~ ) ^l~/"r«\~/ T\^7/~r • • •
8y = _ _
VA;/ 2\*
Dividing all through by SA; —
8y = i 8x (8x)* (Sx)3
8x~ x 2x2 3%3 4** '
By sufficiently diminishing the value of Sx we may make the
52 MATHEMATICS FOR ENGINEERS
second and succeeding terms as small as we please, and evidently
the limiting value of the series is -
i* c,. ,
dx
LSy _ i
8x x
Hence —
Sx—>o
d
_
dx ~ x
(b) Assuming the result — —— = ex
Qt%
Let — y = loge x, so that x — ey
dx dey
and — = - = ey.
dy dy
^>. .
Now ~ =
OA/
T
r- and consequently by considering the limiting
oA/
*y d
values of these fractions — /- = -r-
dx dx
dy
dv
• We wish to find ~ and we have already obtained an expression
, dx
-=-.
dy
Hence — ~ = T- = — . =
dx ax ey x
dy
d loge x _ i
\JL 5 — —
dx~ x
This result can be amplified to embrace the more general form,
s —
thus —
for, in accordance with the rule given on p. 48, the A which
multiplies the I.V. in the original fun6tion must appear as a
multiplier after differentiation.
All these rules apply to functions involving natural logs, but
DIFFERENTIATION OF FUNCTIONS 53
they can be modified to meet the cases in which common logs
occur ; for - Iog10 x = -4343 loge x
d logln x d loee x
and hence -$&- = .4343 _J3£L_ =
d . ,. -4343A
and -- Iog10 (Ax+B) -
It should be observed that in all these logarithmic functions
the I.V. is raised to the first power only : if "the I.V. is raised to
a power higher than the first, other rules, which are given later,
must be employed.
Example 20. — If y = \oge jx, find ~.
dy d .
-r**-T loge 7
dx dx _
or alternatively — loge 7* = loge 7+loge x
and thus— loge 7* =
. i i
= o+- = -
X X
Example 21. — Differentiate with regard to t the expression
Iog10 (5^—14) and find the numerical value of the derivative when
/ = 3-2.
When t = 3- 2
10£T Ut- 1^ - '4343 X 5 -
dt tog], \y • i_r -
log,. (51-14) = = "0858.
We may check this result approximately by taking values of J
3-19 and 3-21 and calculating the value of -' *! -- —•
Thus—
When i = 3-19, Iog10(5f— 14) = Ioglfl(i5'95— 14) = Iog10 1-95 = -2900
when t = 3-21, Iog10(5*— 14) = Iog10(i6-o5— 14) = Iog10 2-05 = -3118
so that —
8 Iog10 (5^—14) = -3118 — -2900 = -0218
while 8t = 3-21 — 3-19 = -02
S . , -0218
and lo <— * ==~~ = I>O9-
54 MATHEMATICS FOR ENGINEERS
Differentiation of the Hyperbolic Functions, sinh x and
cosh x. — Expressing the hyperbolic functions in terms of exponential
functions —
ex— erx
sinh x =
and cosh x =
2
ex-\-e-x
2
Thus to differentiate sinh % we may differentiate
d sinh x d (ex—e~x\ i, _
H£»r»^£» — I — _ / x*g I x> — £\
jncii^c j — 7 i ~ / — lu ~ i c? i
»A; «A;\ 2 / 2
= cosh x
d d(ex+e-x\ i. ,
also j- cosh « = j-( - - ) = -wf—g-'i
dx dx\ 2 / 2V
= sinh #.
Example 22.— Find the inclination to the horizontal of a cable
weighing \ Ib. per ft. and stretched to a tension of 30 Ibs. weight, at
the end of its span of 50 ft.
The equation to the form taken by the cable is —
/ x —x\
y = -\e c-\-e c) = c cosh -
horizontal tension 30 ,.
where c = --- — Cr -••*_— =60.
weight per foot -5
We require the slope of the curve when x = 25, this being given
cl/\}
by the value of - there.
dy d , x i . . x • •> x
~- = J- c cosh =— = c x >— sinh >- = sinh 7-
dx dx 60 60 60 oo
. <
When x — 25 -/ = sinh -^ = sinh -4167 =
g-
4167 _ g— .4167
-
dx 60 2
= 1-517 — 659
2
= -429.
«
This value is that of the tangent of the angle of inclination to
the horizontal; which is thus tan-1 -429 or 23° 13'.
DIFFERENTIATION OF FUNCTIONS 55
Exercises 4. — On the Differentiation of ax, Log x and the Hyperbolic
Functions.
Differentiate with respect to # the functions in Nos. i to 20.
1. e~5*. 2. i-se4-1*. 3. -£. 4. 4-15*. 5. 8-72^.
&
. 7. ^xe. 8. 14x2*. 9. 4ie
foa-iX -jpl-ZX
10. 3-i4«»^-5*»-« + 3-ite+6. 11- °,VX^T*E- 12. log 7*.
13. 3 log (4— 5*). 14. iolog108#. 15. c log (4*0 +56).
16. 9«--te-log -2*+^. 17. log 2*(3#-47).
18. log y)"-- 19- (**)3+4 cosh 2*- 17 loglo 2-3*.
20. log 3*2+5#-'7 — i -8 (i -8*) + 12.
21. If y = A^-^+A^-^find the value of S+7
2 ' '
22. Find -5- log (3 — 4^) when v = 17. Check your result approxi-
CL1)
mately by taking as values of v 1-65 and 175.
23. Determine the value of -,- 71 log (18 — -04^).
(I If-
24. Write down the value of -j- Iog10 18^.
at
25. If T = 5oe'Zd, find the rate of change of T compared with
change in 6.
/ _BA
26. If C = CQ^I— e L '), C and C0 being electrical currents, R the
resistance of a circuit and L its self-inductance, find the rate at which
the current C is changing, / being the time.
dl)
27. Given that v = 2-03 Iog10 (7— i-Su), find ^-.
28. Evaluate -=- 5 cosh - and also -j- p sinh -.
ax 4 ay r q
29. An electromotive force E is given by —
E — A cosh Vlr . #+B sinh Vlr . x.
Find the value of -3-3 in terms of E.
30. If W = i44J/)1(i f log r) — rpb], find the value of r that makes
-T— = o ; W being the work done in the expansion of steam from
pressure pt through a ratio of expansion r.
31. Find the value of -r~ + #y if y — e~ax.
ax ' * a a
56
32. If y = AeZx-\-'Be3x+Ce-*x> find the value of—
d3y dzy dy
dzV Vyx x/n-*, -J.x
33. Evaluate -7-2 -- when V=A1ex^ r2 +A2e ^ r2
W^? ? 2
34. Nernst gives the following rule connecting the pressure p of a
refrigerant (such as Carbon Dioxide or Ammonia) and its absolute
temperature T —
p = A+B log T+Cr+5
T
where A, B, C and D are constants. Find an expression for -J-.
Differentiation of the Trigonometric Functions. — Before
proceeding to establish the rules for the differentiation of sin x
and cos x, it is well to remind ourselves of two trigonometric
relations which are necessary for the proofs of these rules, viz. —
(a) When the angle is small, its sine may be replaced by the
angle itself expressed in radians, i. e. —
T sm6 = T (cf part ^ p 458)>
(b) sin A-sin B = 2 cos sin - (cf. part I, p. 285).
\ 2 / 2
To find -v- sin x we proceed as in former cases ; thus —
Let y = sin # and y+8y = sin (#-f 8#)
then 8y — y -\-8y-y = sin (x-\-8x)—sin x
(2x+8x\
= 2 cos ( - sin
\ 2 /
Dividing through by 8x —
_ 2 COS \ ~~ / OAII i
8y V 2 / \2
/2^+8^\ . /Bx\
I sin (— J
8* 8x
(2x+%x\ f8x^
COS ( • I sin I
t\ (8X\
- ) sm ( — I
/ \2/
DIFFERENTIATION OF FUNCTIONS 57
The limiting value of ?- is -£-. and that of the right-hand side
is cos x, since cos ( x-\ — J approaches more and more nearly to
cos x as 8* is made smaller and smaller, and the limiting value of
/sin — \
/ 2 \ .,,.,., sin 9 .
I, or, as we might write it, — ^— , is I.
Hence
d sin x dy
—j — or * =
dx dx
L
-^ = cos x
dsinx
dx
= cos x.
•75
-5
•25
O
-y
-y-
.JG
«-»
7
.3G
TT.
x
JG
FIG. 14. — Curves of y = sin x and y •= cos x.
By similar reasoning the derivative of cos x may be obtained ;
its value being given by —
The graphs of the sine and cosine curves assist towards the
full appreciation of these results. In Fig. 14 the two curves are
plotted, and it is noted that the cosine curve is simply the sine
curve shifted backwards along the horizontal axis : thus the slope
curve and the primitive have exactly the same shape. This
condition also holds for the primitive curve y = e?1, and so suggests
that there must be some connection between these various natural
functions; and further reference to this subject is made later in
the book.
58 MATHEMATICS FOR ENGINEERS
Much trouble is caused by the presence of the minus sign in
d COS OC
the relation — -5 — = — sin x, it being rather difficult to remember
whether the minus sign occurs when differentiating sin x or cos x.
A mental picture of the curves, or the curves themselves, may be
used as an aid in this respect. The cosine and sine curves differ
in phase by £ period (see Fig. 14), but are otherwise identical.
Treating y = sin x as the primitive : when x is small, sin x and x
are very nearly alike, and thus the slope of the curve here is i;
as x increases from o to - the slope of the curve continually
7T
diminishes until at x = - the slope of the curve is zero. Now
the ordinate of the cosine curve when x = o is unity, and it
diminishes until at x = ~ it is zero. From x = ~ to x = -n- the
2 2
slope of the sine curve is negative, but increases numerically to — I,
this being the value when x = TT; and it may be observed that
the ordinates of the cosine curve give these changes exactly, both
as regards magnitude and sign. Thus the cosine curve is the
slope curve of the sine curve.
Now regard the cosine curve as the primitive. At x = o the
curve is horizontal and the slope = o ; from x = o to x = ~ the
2
slope increases numerically, but is negative, reaching its maximum
negative value, viz., — i, at x = - ; but the ordinates of the sine
2
curve are all positive from x = o to x = ~, so that although these
2
ordinates give the slope of the curve as regards magnitude, they
give the wrong sign. In other words, the sine curve must be
folded over the axis of x to be the slope curve of the cosine
curve, i. e., the curve y = — sin x is the slope curve of the curve
y = COS X.
To summarise, we can say that the derived curve for the sine
curve or for the cosine curve is the curve itself shifted back along
the axis a horizontal distance equal to one-quarter of the period.
Thus we can say at once that the slope curve of the curve
y = sin (x-\-b) is the curve y — cos (.r+fr), since the curve
y = sin (x+b) is the simple sine curve shifted along the horizontal
axis an amount given by the value of b, the amplitude and period
being unaltered.
DIFFERENTIATION OF FUNCTIONS 59
Thus— -,- sin (x+b) = cos (x+b)
and, in like manner —
jx cos (x+b) = - sin (x+b).
Again, -7- sin (5#+6) = 5 cos (5#+6), since 5 multiplies the
I.V. in the original function.
Then in general —
?- A sin (Bx+C) = AB cos (Bx+C)
-? A cos (Bx+C) = — AB sin (Bx+C).
To differentiate tan x with regard to x.
Let y = tan x and (y+8y) then = tan (x-\-8x)
8y = y+Sy— y=tan (x+8x) — tan x
_ sin (x-\-8x) sin #
~ cos (x+8x) cos A;
_ sin (x+Sx) cos A;— cos (*+SA;) sin x
cos #8# cos x
sin
cos (x+8x) cos x
_ sin 8x
~ cos (A; +8x) cos x
Dividing through by 8x —
Sy _ sin 8x I
8x ~ 8x cos (x+8x) cos x
Now as 8x approaches zero, - - approaches i and (x + 8x)
approaches x.
dx ,8* * ^ cos A; cos x cos2 A;
Hence — =1 = IX- ~= *~ = sec2 x.
\ _ ,
3- tan x = sec2 x
60 MATHEMATICS FOR ENGINEERS
In like manner it can be proved that —
d cot x
— 3 — = — cosec2 x
d seex sinx
dx ~ cos2 x
d cosec x cos x
and — -j~ = -- r-s —
dx sin2 x
To generalise —
^ A tan (Bx+C) = AB sec2 (Bx+C)
J^ A cot (Bx+C) = — AB cosec2 (Bx+C)
rf , ,_ _v AB cos (Bx+C)
a- A eosee (BX+C) = - .2
AB
.
cos2 (Bx+C)
Example 23. — Find the slope of the curve representing the equation
s = 5-2 sin (40^—2-4) when t — -07.
The slope of the curve —
. . .
5'2 Sm (4°^— 2'4)
= 208 cos (40^—2-4).
ds d . . .
= ~dt = dt5'2 Sm (4°^— 2'4) = 5-2x40 cos (40^—2-4)
Hence when —
t = -07, the slope = 208 cos (2-8—2-4) = 208 cos -4 (radian)
= 208 cos 22-9°
= 192.
Example 24. — Differentiate, with regard to z, the function
9-4 cot (7—5?).
,- 9'4 cot (7-5*) = 9'4 X -5 X - cosec2 (7-5?)
U»6
= 47 cosec2 (7—52).
Simple Harmonic Motion. — We can now make a more strict
examination of simple harmonic motion. Suppose a crank of
length r (see Fig. 15), starting from the position OX, rotates at
a constant angular velocity <o in a right-handed direction. Let it
have reached the position OA after t seconds have elapsed from
the start; then the angle passed through in this interval of time
= AOM = <&t, since the angular distance covered in I sec. = o>
radians and the angular distance in t seconds = a>t radians.
DIFFERENTIATION OF FUNCTIONS
61
Considering the displacement along the horizontal axis, the dis-
placement in time t = s = OM
= AO cos AOM = r cos tat.
ds
Then the velocity = , , = — rXta sin tat = — rta sin <at
and the acceleration = ,, = —
at
cos tat = — tazxr cos wt
= — 0)2S
*'. e., the acceleration is proportional to the displacement, but is
directed towards the centre : thus, when the displacement from
the centre increases, the acceleration towards the centre increases.
When the displacement is greatest, the
acceleration is greatest : e. g., if the
crank is in the position OX, the
acceleration has its maximum value
wV and is directed towards the centre,
just destroying the outward velocity,
which at X is zero. At O the acceler-
ation = — u>2x o = o, or the velocity is
here a maximum.
An initial lag or lead of the crank
does not affect the truth of the foregoing FlG r .
connection between acceleration and
displacement. The equation of the motion is now s = r cos
where c is the angle of lag or lead, and the differentiation to find the
velocity and the acceleration is as before.
Example 25. — If s — 5 sin 4* — 12 cos \t, show that this is the
equation of a S.H.M. and find the angular velocity.
s = 5 sin ift — 12 cos 4^.
Then — v = ,$ = (5 x 4 cos 4*) — (12 x 4 X — sin 4*).
HP
= 20 cos 4^+48 sin 4^
and a = " = (20 x 4 X — sin 4*) + (48 x 4 cos 4/)
= — 80 sin 4^+192 cos 4*
= —16(5 sin 4^—12 cos 4/) = — i6s
i. e., the acceleration is proportional to the displacement.
Now, in S.H.M., the acceleration = — «2s.
&>2 = 16, i. e., w = angular velocity
= 4 radians per sec.
62 MATHEMATICS FOR ENGINEERS
This last question might be treated rather differently by first
expressing 5 sin 4^—12 cos 4^ in the form Msin(4^+c) (see
Part I, p. 276) and then differentiating. This method indicates
that a S.H.M. may be composed of two simple harmonic motions
differing in phase and amplitude.
Exercises 5. — On the Differentiation of Trigonometric Functions.
Differentiate with respect to x the functions in Nos. i to 16.
1. sin (4— 5'3x). 2. 3-2 cos 5-1*. 3. -16 tan (3X-\-g).
4. 2-15 sin i1 ^~5). 5. 8 cot 5%.
\ 4 /
6. 43-15 sec (-05 — -117*). 7. be cos (d—gx).
8. 4 cos $x— 7 sin (2^—5). 9. sin 5-2^ cos 3~6x
10. 2-17 cos 4-5* cos 1-7*. 11. 9-04 sin (px+c) sin (qx—c).
12. 5 sin2 x. 13. -065 cos2 3*.
14. cos2 (7*— i-5)+sin2 (7*— 1-5).
15. 3*1-72— 5- 14 log (3#-4-i) + -i4 sin (4-31 — -195*) + 24-93*.
16. 7-05 sin -015*— -23 cos (6-i — -23*) + 1-85 tan (4*— -07).
17. x, the displacement of a valve from its central position, is given
approximately by x = — 1-2 cos a/— 1-8 sin «^ where w = angular
velocity of crank shaft (making 300 r.p.m.) and t is time in seconds
from dead centre position.
Find expressions for the velocity and acceleration of the valve.
18. If 5 = 4-2 sin (2-1 — -172) — -315 cos (2-1 — -J7/), s being a displace-
ment and t a time, find an expression for the acceleration in terms of
s. What kind of motion does this equation represent ?
19. The current in a circuit is varying according to the law
C = 3-16 sin (2irft— 3-06). At what rate is the current changing when
t — -017, the frequency / being 60 ?
20. If the deflected form of a strut is a sine curve, what will be
the form of the bending moment curve ?
21. If y = deflection of a rod at a distance x from the end, the
end load applied being F —
Bl
y=-
8COS1
Find the value of EI~^-}-Fy-\--& cos -j-; y and x being the only
variables.
22. The primary E.M.F. of a certain transformer was given by the
expression —
E = 1500 sin pt-\-ioo sin 3^—42 cos ^+28 cos 3pt.
Find the rate at which the E.M.F. varied.
T 2
23. A displacement s is given by s = sin izt sin 13^. Show
that the acceleration = 25 sin 12^—1695.
CHAPTER III
ADDITIONAL RULES OF DD7FERENTIATION
Differentiation of a Function of a Function. — \Vhilst the
expression e sin 4* is essentially a function of x, it can also be spoken
of as a function of sin 4*, which in turn is a function of x; and
thus it is observed that e aa ** is a function of a function of x.
This fact will be seen more clearly, perhaps, if u is written in place
of sin 4* : thus e sin ** = ew, which is a function of u, which,
again, is a function of x, since u = sin 4*.
To differentiate a function of a function the following rule is
employed —
dy_=dyxdu
dx du dx
\
and this rule is easily proved.
Let y be a function of u, and let u be a function of x : then y
is a function of a function of x. Now increase # by a small
amount 8x; then since u depends on x, it takes a new value
u + 8u, and also the new value of y becomes y + Sy. Since these
changes are measurable quantities, although small, the ordinary
rules of arithmetic can be applied, so that —
8y __Sy 8u_
Sx~SuXSx
When Sx approaches zero these fractions approach the limiting
values -, and - respectively : and thus in the limit —
ax du dx
dy _ dy du
dx du dx
In like manner, if y is a function of u, u a function of w, and
w a function of x, it can be proved that —
dy_dy..du_dw
dx~duxdwxdx
63
64 MATHEMATICS FOR ENGINEERS
It will be observed that on the right-hand side of the equation
we have dy as the first numerator and dx as the last denominator
(these giving in conjunction the left-hand side of the equation) ;
and we may regard the other numerators and denominators as
neutralising one another. The simple arithmetic analogy may help
to impress the rule upon the memory : thus —
Example i . — If y = e 8to tx, find the value of ~ .
dit
Let u = sin 4*, so that j- = 4 cos 4* and y = eu.
Since y is now a function of n, we can differentiate it with regard
to u, whereas it is impossible to differentiate with regard to x directly.
y = eu and ~ = ~ = eu = e s^ **.
du du
T,, dy dy du
Then, since -, - = -/-x j-
dx du dx
dv
JL — e sui tx x ^ cos 4#
= 4 cos *g sin
Example 2. — Find the value of j~log (cos 2x}3.
Let v = (cos 2x)3 and u = cos 2x; and thus y = loge v and u = w3.
dy dy dv du u = cos
Then -~ = -~ x -j- X -j-
d log v du3 d cos 2.x
~~dv~X~duX dx
X — 2 sin
-j- — — 2 sin
V = U3
dv
-j- = 3U
du °
_ — 6 sin 2XX (cos 2^r)2 _ — 6 sin 2
(cos 2#)3 cos 2x
= —6 tan 2
Example 3. — The radius of a sphere is being decreased at the rate
of -02 in. per min. At what rate is (a) the surface, (b) the weight,
varying, when the radius is 15 ins. and the material weighs -3 Ib. per
cu. in. ?
dr
If r — radius, then - ,- = rate of change of the radius, and is in
dt
this case equal to —-02.
ADDITIONAL RULES OF DIFFERENTIATION 65
(a) The surface = 4irf2, and thus the rate of change of surface
_ dS
~ dt
_ d . 4*rz
dt
dr*
= *v'^dt
dr* dr
= *v-~drXdf
dr
5
—'02.
Hence when r = 15, -5- = 8v x 15 X — -02 = —7-53, i. e., the surface
is being diminished at the rate of 7-53 sq. ins, per min.
(b) The volume = 4 ^
so that the rate of change of volume = — = jft-***)
and the rate of change of the weight = -rr = -j-.( - X -Sirr3 j
<AV d dr3 dr3 dr
~rr =~JT- '4*r — '4* • -JT = '4* X -j- X T-.
—-02.
When r = 15, -, = -4^x3x225 x —-02 = —16-93
or the weight is decreasing at the rate of 16-93 IDS- per min.
Example 4. — Find expressions for the velocity and acceleration of
the piston of a horizontal steam engine when the crank makes n
revolutions per second.
In each turn the angle swept out = 2ir radians.
Hence in i second 2vn radians are swept out, i. e., the angular
velocity = 2.im; and this is the rate of change of angle, so that
dQ
dT=2vn-
From Fig. 16 CD = / sin a
and CD = r sin 6.
Thus I sin a = r sin 6
/ r
or sin 0 = - sin a, and sin a = T sin 0.
r I
66
MATHEMATICS FOR ENGINEERS
Again, cos « = Vi — sin2 a
If the connecting rod is long compared with the crank, -j is small
rz
and 72 still smaller, so that our method of approximation can be
p
applied to the expansion of the bracket, i. e. —
I y^
cos o = i 72 sin2 0, very nearly.
FIG. 16. — Velocity and Acceleration of Piston.
Let AB = displacement of the piston from its in-dead-centre position
= x = AE+OE— BO = l + r-BV — DO
= l+r—l cos o— r cos 6
= l+r—l (i — £p sin2 0)-y cos 0
— y+~, sin2 0— Y cos 0
= r-\ — —
dx d
i — cos 20
2
r2 cos 20
-r cos 0
fl
~ — r COS U.
r2 r2 cos 20
XT . i. i • . ,,-. • . dx a f
JNow the velocity of the piston = -=- = --=•- -j , --- -.
We cannot differentiate this expression directly, so we writ
n
r cos 0
TT
Hence
dx __df r*
dx dx dQ
• dQ
COS 20
~dt = 50 X dt'
,.) dQ
rcosQ\x,.
J a/
= -jo-f o — (~ x —2 sin 20J — (rx —sin 0) j- x 2irW
fy sin 20 . . 01
= 2trwy -j - j -- f-sm 0 j-
or if
- = m
r
dx
V = ~dt =
f sin 20
\ 2m
ADDITIONAL RULES OF DIFFERENTIATION 67
dv dv dQ d (sin 20 , _i d0
Also the acceleration = -rr = -j^X^—-^,. zvnrl --- f-sm 6 [ x -r-
at ay at at) I 2m at
fcos 26 , n)
= 2irnr 1 -- 1- COS 6 h X 2irW
„ „ f Q . COS 26"|
= 4ir2n2»' { cos 0H —
* I »» /
Example 5. — Water is flowing into a large tank at the rate of
200 gallons per min. The reservoir is in the form of a frustum of a
pyramid, the length of the top being 40 ft. and width 28 ft., and the
corresponding dimensions of the base being 20 ft. and 14 ft. ; the
depth is 12 ft. (see Fig. 17). At what rate is the level of the water
rising when the depth of water is 4 ft. ?
In 12 ft. the length decreases by 20 ft., and therefore in 8 ft. the
length decreases by — , i. e., 13 J ft., so that the length when the
water is 4 ft. deep is 40— 13^ = 26| ft.
Similarly, the breadth = 28— (f x 14) = i8f ft.
i. e., the area of surface = 26f x i8| = 498 sq. ft.
200 ,,
200 gals, per mm. = ^ — cu. ft. per mm.
0*24
= 32-1 cu.. ft. per min.
i. e., the rate of change of volume = —rr = 32-1.
dv
Now -v- = -v . Ah, where A = area of surface
at at
and h = depth of water,
*dh f since for the short interval of time considered the\
dt \ area of the surf ace may be considered constant./
Hence the rate of change of level = -JT = -jr X -z-
dt dt A
32-1 xi , ,.
= — — Q— = -0644 ft. per nun.
49°
= »773 in. per min.
68 MATHEMATICS FOR ENGINEERS
Example 6. — If a curve of velocity be plotted to a base of space,
prove that the sub-normal of this curve represents the acceleration.
d'v
The sub-normal of a curve = y~- (see p. 43).
CbX
In this case, since v is plotted along the vertical axis and s along
the horizontal axis —
the sub-normal = v-^-
ds «.
dv dt
V'dtXds
= vxax--
v
= a
[for -=r = rate of change of velocity = a\
dt
ds
dt
and -3T = rate of change of space = v I
As a further example of this rule, consider the case of motion due
to gravity; in this instance v2 — 2gs, i.e., the velocity space curve
is a parabola. Hence we know that the sub-normal must be a
constant, i. e., the acceleration must be constant.
The sub-normal = v-r
ds
,T dv2 d ds
Now ds=ds-^S = ^-ds = ^
dvz dvz dv dv
but ,— = -j— . -T- = 2v j-
ds dv ds ds
dv
2V-j- = 2g
ds
dv
ds
i. e., the sub-normal or the acceleration = g.
Vds =
Exercises 6. — On the Differentiation of a Function of a Function.
T-" j A d sin 2x d . „ 0 d
Find 1. -3- . e . 2. -T- log v2. 3. ^2cos2 t.
dx dv dt
d _ d n d sin 5x
dx Sm dx 3'14 (5^2+7^~2)- 6. ^a
7< ^1<88' 8< dx logl° (3 + 7^-9^3). 9. ^ cos (log s5).
and 10. ~ log tan -.
a* 2
ADDITIONAL RULES OF DIFFERENTIATION 69
11. In the consideration of the theory of Hooke's coupling it is
required to find an expression for — , i. e., a ratio of angular velocities.
If o>B = §r, <"A = -3T and tan <f> = - — , find an expression for
— B in terms of the ratios of 9, </> and o.
"A
12. Find an expression for the slope of the cycloid at any point.
The equation of the cycloid is x = a (6 + sin 6)
y — a(i — cos 6)
the co-ordinates # and y being measured as indicated in Fig. 18.
^Rolling Circle
FIG. i 8.
13. Assuming that the loss of head due to turbulent flow of water
in a pipe is expressed by h — C(AV2+BV?), where V = mean velocity
of flow in ft. per sec. ; show that the slope of the curve in which log h
and log V are plotted with rectangular co-ordinates is given by —
d log h
dlogV
2A
14. If 3x*+8xy+5y2 = i
show that T = T
dxz (.
15. A vessel in the form of a right circular cone whose height is
7 ft. and diameter of its base 6 ft., placed with its axis vertical and
vertex downwards, is being filled with water at the rate of 10 cu. ft.
per min. ; find the velocity with which the surface is rising (a) when
the depth of the water is 4 ft. and (b) when 60 cu. ft. have been
poured in.
16. If p = (r)K, prove that-j^ = ~ffi(— r^ 1°§ r-
17. If x3— 6x2y— 6xyz+y3 — constant, prove that —
dy _ xz—4xy—r--*
dx= 2X*~
7o MATHEMATICS FOR ENGINEERS
18. A ring weight is being turned in a lathe. It is required to
find the weight removed by taking a cut of depth ^thj". The material
is cast iron (-26 Ib. per cu. in.), the outside diameter of the ring is
3-26" and the length is 2-5'*. Find the weight removed.
Find also a general expression for the weight removed for a cut
of depth ^J^" at any diameter.
19. Find the value of -rA log tan —
20. If P = -^TT, and -„.., = u, find -w. (This question has refer-
-„..,
ence to stresses in redundant frames.)
21. Find the angle which the tangent to the ellipse — \-— = 2 at
4 9
the point x = 2, y — —3, makes with the axis of x.
22. Find the slope of the curve 4#2+4y2 = 25 at the point x = 2,
y = — f, giving the angle correct to the nearest minute.
23. If force can be defined as the space-rate of change of kinetic
, , . , . wvz , , wa
energy, and kinetic energy == - , prove that force = — .
o o
dx
24. If x — 8 log (i2t3— 74), find the value of ,-,.
ctt
Differentiation of a Product of Functions of x. — It has
already been seen that to differentiate the sum of a number of
terms we differentiate the terms separately and add the results.
We might therefore be led to suppose that the differentiation of
a product might be effected by a somewhat similar plan, viz., by
multiplication together of the derivatives of the separate factors.
This is, however, not the correct procedure ; thus —
d ,, „, , . d log x dxz . i
j- (log xxx2) does not equal , --X-T— , ^. e.,— X2x or 2.
dx v dx dx x
The true rule is expressed in the following manner : If u and v
are both functions of x, and y = uv, i. e., their product —
dy d , . du . dv
B -«*"*-«£+•«
Proof. — Let x increase by an amount 8x; then since both u
and v are dependent on x, u changes to a new value w+Sw and
v becomes v -f- 8v.
Now y = uv, and hence the new value of y, which can be
written y-\-8y, is given by —
but
y = uv
ADDITIONAL RULES OF DIFFERENTIATION 71
whence by subtraction —
Sy = y+8y— y = (u-}-8u)(v-}-8v)—uv
= uv-}- u8v -\-v8u-\-8u . 8v — uv
{-8u . 8v.
Dividing through by 8x —
8y_ 8v <^ , s 8v
Sx ~ U^x^rV^cr U ' 8x
As 8x is decreased without limit, ~, -=- and - approach the
8x 8x 8x ^^
values -/ , -y- and -3- respectively, and the term 8u . .— becomes
dx dx dx 8x
negligible ; so that in the limit —
dy _ du dv
dx dx dx
The rule may be extended to apply to the case of a product
of more than two functions of x. Thus if u, v and w are each
functions of x —
dluvw) d(wV) , ,, . ,
v, — ' = -j~, where V is wntten for uv
dx dx
- wdV^
.j dw
dx^
V dx
nnd(uv)
dw
dx
bUVdx
( du . dv\ .
= w( v-j — \-u-r- )+i
\ dx dxj '
dw
wU "^
dx
and thus —
d(uv
w\ du.
' mil i
dv
du
itit
Example 7. — Find - when y — xz . log x.
T i ,
Let — u = x* so that -r-= zx
dx
and let v = log x so that -j- = -.
(Kx X
™, d .uv du dv .. . , / „ i \
Then —= — = v^- +Uj~ = (log xx 2x) + \ xz . - I
dx dx dx ^ { '^v xJ
= x(l+2 log X).
72 MATHEMATICS FOR ENGINEERS
Example 8. — Find the value of -7,[5e~7' . sin (6^ — 4)]
ctt
cLi>t
Let u = ^e-~l so that — = 5x —je-'1 — — 35e~7t
and let v = sin (6/— 4) so that -,7 = 6 cos (6^—4).
ttt
d . uv du , df
-3T = "-^+M-^
= [sin (6/-4)x -35*-7<] + [5*-7' x6 cos (6/-4)]
= 5g~7*[6 cos (6^—4) — 7 sin (6^—4)].
Example g. — If 2q+~ (pxz) — o, show that 2q = —2p — xf
x dx ax
p being a function of x. This example has reference to thick spherical
shells.
If p is a function of x, px* is of the form uv, where u = p and
v = x2.
d 9dp , dx2 9dp ,
Hence — -,- . px2 = x2,"-\-p-r- = x2-/-4-2Xp.
dx ' r
Hence — 2q-\ --- 3- . pxz — zq+x ,-
1 x dx ' dx
dp
i.e., o = 2q+x ,r
Q//V
Example 10. — Find the value of -y- gx* sin (3^—7) log (i — 5#).
Let u = x*, v = sin (3^—7) and w = log (1 — 5*)
,, du „ dv dw —5 5
then -,~ = 4^3, -; = 3 cos (3^—7) and -y- = — = — —
>V9** sin (3*- 7) log (1-5*) = 9^ ' '
^
F rfw , dv . dw~]
= g{ wv .,--\-wu, -\-uVj
L dx dx dx-1
— 5x) sin (3^-7)4Ar3+{log (i — 5*)*4X3cos (3^—7)}
sin (3*— 7)
= 9*3[_4 sin (3Af— 7) log (1 — 5*) + 3* cos (31*— 7) log (1 — 5*)
5#sin (3*— 7)1
_ + 5*^T~
ADDITIONAL RULES OF DIFFERENTIATION 73
Exercises 7.— On the Differentiation of a Product.
Differentiate, with respect to x, the functions in Nos. i to 12.
1. x2 sin 3*. 2. log 5#X 2#3-4. 3. e9* Iog10 gx.
4. 4Ar-5.tan (3-1 — 2-07*). 5. cos 3-2* cos
6. cos (5 — 3*) tan 2#. 7. 8*1-6 cos
8. 9\ogx3.53*. 9. e*i°g*. 10. ^*
11. 6*te+*(5*+2)«. 12. 7-2 tan ~ log *7.
o
13. If y = Ae3* cos (— + B), find the value of—
14. Find the value of ^/~5< cosh (— 5/).
15. y = (A + B*)*-1*; find the value of ^+8
W^ W
16. If V = 250 sin (jt— -116), A = 7-2 sin 7* and W = VA, find
, d\V
the value of —rr.
at
*17. Differentiate with respect to t the function i^t2 sin (4— -8tf).
18. Find the value of -r,(4*3'7 cos 3/).
•H
Differentiation of a Quotient. — If u and y are both functions
of x, and y = -, then —
dy Vdx udx
dx i>2
Proof.
(a) From first principles. — Let y = - : then a change Sx in x
causes changes of Sy in y, Su in u, and Sy in y, so that the new
value of y = y+Sy = -^pr-.
§u u uv-\-v8u — «y — uSv
™,
Then — Sy = y+Sy— y =
and, dividing through by Sx —
Sy i
. .
y(y+Sy)
_
' Sx~U ' Sx
74 MATHEMATICS FOR ENGINEERS
When 8x becomes very small, P-, .- and ^- approach the values
8x 8x 8x
//-Af CL'IA/ Ul)
-J-, -j- and -j- respectively, whilst v-{-8v becomes indistinguishable
from v.
TT . ,, ,. ., dy i ( du dv\
Hence in the limit -r- = (v . -* — u . -r-
dx vxv\ ax ax/
du dv
V-j tt-j-
dx ax
V2
(b) Using the rules for a product and a function of a function.
u
y = - =
v
TM dy d , ,, ,du , dir1
Then~" = ' (UV ^ = V +U ' ~
x dx
U-v (( -v
I du\ dy-1 dv
du\ . ( 9 dv
_ I If V Tit — « V
. j / T^ I M X\ At/ /\ j
.v a*/ \ ^
du dv
V ~j W ~j~~
dx dx
Example n. — Differentiate, with regard to 5, the expression —
5 cos (35+4)'
T , ,-, du
.Let — w = 45^+75, tnen 3- = i2s^ + 7»
as
dv
and let v = 5 cos (35+4), then -,- = — 15 sin (35 + 4).
ivS
«ZM tiy
„,, d (u\ ds d
Then -.- . I - ) = -
ds \v/ vz
= [5 cos (35+4) X (i252+7)]-[(4S3 + 7*) X -i5sin(35+4)]
25 cos2 (35+4)
5 cos2 (3^+4)
Example 12. — If y — 94*X, -- -, find the value of -£-
Let
u = g4*, then j- = 4 x 94*loge 9 = 4 X 2-1972 x g*
= 8-789 X94z
and let
v = log 7*, then -=- = -*- = -.
du dv
dy d (u\ dx
Hence -/- = T-V - ) = — — ,
dx dx\v I vz
— ui—
dx
(log yx x 8 -79 x 94z) — (g4* X ^
(log 7#)2
94^{(8-79^xlog7^)-i}
(log
FIG. 19. — Spring loaded Governor.
Example 13. — For a spring loaded governor (see Fig. 19)
where Q = force to elongate the spring i unit, T = tension in
spring, W = weight of i ball, « = angular velocity, r — radius of
path of balls, / = length of each of the 4 arms.
If W = 3, g = 32-2 and -j- — 80 when o> = 26, r = -25 and / = i,
find T and Q.
As there are two unknowns, we must form two equations. By
simple substitution —
_ 32'2{T+2Q(i- V 1—0625)}
~"
V 1—0625
- "968
whence T+-o64Q = 60-96 ....... (i)
We are told that -^ must equal 80.
dta
-vr
NOW
Also
-5- = -,- X j- = 2w T-
dr da> dr dr
(2)
d
dr
where
and
— r* dr\v
u = g{T+2Q(l— Vl*—rz)}
v =
76 MATHEMATICS FOR ENGINEERS
Thus to determine -r- and -y- it is first necessary to find the value
rJ A//2 4,2
of : to do this let l*-r* = y
dr
then
Thus
and
du dv
, VT U^
„,, aw dr dr
Then - - = » —
so that ' — — zr
dr
/.,\
Thus, differentiating both sides of the original equation with respect
to r, we have from (2) and (3) —
2oi T- = ° ,
rfy W
Substituting the numerical values —
2x26x80 =
52X ;° ^9375 X- 968 _ .^SQ+T+^Q . 2g+T
whence J4Q7 = 2Q+T
but from (i) 60-96 = -064(3 +T
and therefore Q — 695-3!
and T = 16-4!
Differentiation of Inverse Trigonometric Functions. —
Since inverse trigonometric functions occur frequently in the study
of the Integral Calculus, it is necessary to demonstrate the rules
for their differentiation; and in view of their importance in the
later stages of the work, the results now to be deduced should be
carefully studied.
The meaning of an inverse trigonometric function has already
been explained (see Part I, p. 297), so that a reminder only is
ADDITIONAL RULES OF DIFFERENTIATION 77
needed here. Thus sin~x x is an inverse trigonometric function,
and it is such a function that if y = sin"1 x, then sin y = x.
To differentiate sin~l x with regard to x.
Let y = sin-1 x so that, from definition, sin y = x
then
but
and hence
or
milarlv —
dx ~ dx~
d sin y d sin y dy
dx dy dx
dy
i = cosyx^
dy T. i *
d# ~~ cos y Vi— sin2 y ~ Vi—x2
d »in 1 y 1
-?— sin jc — / —
dx Vi y2
d 1
^— COS 1 Y — — /
(x being supposed to vary between o and -).
Example 14. — Find the value of -5- tan-1 -.
Let
y =
tan-1 -,
a
*'. e.,
tan y
and
Now
but
sec2 y =
d tan y
i+tan*
d (x\
i
a2
a2
d#
d tan y
dx \a'
d tan y
a
vrfy
d*
dy
a^
Hence
i
a
sec2yx
S
tfy
i
i
a2
d# a sec2 y a a2 +#2
a
78 MATHEMATICS FOR ENGINEERS
Example 15. — Find the value of ~ cosh"1 -.
ax a,
Let
y
, - X
= cosh"1
a
then
cosh y
a'
So fli'if
d cosh y
d (x\ i
dx
d!# \a) a
i nit
d cosh y
d cosh y dy
rf#
dy Xdx
hence
i
= sinh y X ^
a
•^ rf*
(i)
Now cosh2 y— sinh2 y = i
Xs
whence sinh2 y = cosh2 y— i = -g— i
a*
and sinh y = ±- \/#2— a8
Then, substituting this value for sinh y in (i) —
2 y
a x
dy , i
or -/ = ± •>-=-
d , «x .
v- cosh"1 — = ±
„
2— 2
Exercises 8.— On the Differentiation of a Quotient and the
Differentiation of Inverse Functions.
Differentiate with respect to x the functions in Nos. i to 12.
1 5^1 2 log (2~7*)
elx~5' cos (2— 7*)'
3. . ^X » _. O^f
. 5 sin — . Tf* cos — ji.
7 rf2
_ 52-to _ cosh 1-8^
&- g9^i' 4l-8a: -•
I *+3 7 cos-1 3^
7- VT'^W- 8' Vf^^2'
q ^o(a x)x ._ ^
a' 2(6-* cot B)* 1U> a2(a2+Ar2)i'
.. /3 — 6lzx -j- 1 2/^r2 — 7*3 (an expression occurring in the solution of
"• 3^—4-^ a b63-111 problem).
12.
ADDITIONAL RULES OF DIFFERENTIATION 79
e sin (l-ar+1-7)
log (8**-7*+3)'
13. Assuming the results for T- cosh # and T- sinh #, find the value
of -j- tanh #.
d#
Nos. 14 and 15 refer to the flow of water through circular pipes;
v being the velocity of flow, Q the quantity flowing, and 0 being the
angle at the centre subtended by the wetted perimeter.
i T* I sm
14. If, = 13-1(1— Q
sin 26 ^0
17. If » (a velocity) = r* (sin 6+Z and = "' find
15. Given that Q - 132-4 l. find <.
0s «<*
16. Differentiate, with respect to y, the expression —
'-tan-iy.
2
sin 26
di
the acceleration (-57 ); find also the acceleration when 0 is very small.
40 -it • sin 0 , dQ - , , , , ., (d<i>\ ,
18. If sm <t> — -- , and -j-. = a>, find the angular velocity I -if) of
m at J \dt J
a connecting-rod and also the angular acceleration / •— j.
19. Given that ^ = TT - . . f find J-R and hence the value of
(p—q) tanO a0
tan 0 that makes -^ = o.
ao
o/> rs- j A*. j; <^M ,, WX(l— X)(l — 2X) ,, .
20. Find the value of -=— when M = — , , — ? — -. M is a
dx 2(3/— 2x)
bending moment, I is the length of a beam and x is a portion of that
length.
21. Differentiate, with respect to /, the quotient — -— ;,- -- -.
Partial Differentiation. — When dealing with the equation
PV = CT in connection with the theory of heat engines, we know
that C alone is a constant, P, V and T being variables. If one of
these variables has a definite value, the individual values of the
others are not thereby determined ; e. g., assuming that C and T
are known, then so also is the product PV, but not the individual
values of P and V. If, now, the value of one of these is fixed,
say of P, then the value of V can be calculated : therefore V
depends on both P and T, and any change in V may be due to a
change in either or both of the other variables. To find the change
in the value of V consequent on changes in values of P and T,
8o
MATHEMATICS FOR ENGINEERS
the change in V due to the change in P (assuming that T is kept
constant) is added to the change in V due to the change in T
(P being kept constant). Rates of change found according to this
plan are spoken of as partial rates of change, or more usually
partial derivatives, and the process of determining them is known
as partial differentiation.
When only two variables occur, a plane curve may be plotted
to depict the connection between them, but for three variables a
surface is needed. The three co-ordinate axes will be mutually at
FIG. 20.
right angles, two in the plane of the paper, and the other at right
angles to it. If x, y and z are the variables, we can say that z is
a function of x and y, or, in the abbreviated form z = f(x, y).
Similarly — x = f(y, z)
and y = f(x, z).
Dealing with the first of these forms, and assuming the axes
of x and y to be horizontal (Fig. 20), let us examine, from the
aspect of the graph, the significance of this form. Giving any
value to x, we know the distance of the point in front of or behind
the paper : the value of y determines the distance to the right or
left of the axis of z, i. e.t the vertical on which the point lies is
ADDITIONAL RULES OF DIFFERENTIATION 81
determined and the actual height up this vertical is fixed by the
value of z. If z is kept constant whilst values of x and y are
chosen, a number of points are found all lying on a horizontal
plane, and if all such points are joined we have what is known as
a contour line. Therefore, if one of the quantities is constant our
work is confined to one plane; but we have already seen that
when dealing with a plane, the rate of change of one quantity
with regard to another is measured by the slope of a curve, hence
we can ascribe a meaning to a partial derivative.
To illustrate by reference to a diagram (Fig. 20).
The point P on the surface is fixed by its co-ordinates x, y
and z, or SQ, OS and QP.
If x is kept constant, the point must lie on the plane LTND.
The slope of the curve LPT, as given by the tangent of the angle
PMN, must measure the rate of change of z with regard to y when
x is constant; and this is what we have termed the partial
derivative of z with regard to y. This partial derivative may be
expressed by «-, or, more conveniently, by ( -j- ) , and if there is
no possibility of ambiguity as to the quantity kept constant the
suffix x may be dispensed with.
fdz\ nn TXT (the slope being negative, since z
(-=-)= —tan L PMN v
\ay] decreases as y increases).
Similarly, the slope of the curve KPH
_ /«fe\
\dx)'
If the variables are connected by an equation, the partial
derivatives can be obtained by the use of the ordinary rules of
differentiation.
Example 16. — Given that z —
\ (dz
)' (dy
dz\ ldzz
)> W
To find (j ), i. e., to find the rate of change of z with regard to x
when y is constant, differentiate in the ordinary way, but treating y
as a constant.
Thus — K- = (5>> x 2x) — (zyz x 3**) + 2oyexy
= loxy — 6 r2y2 + 2oyery
and yj = (loy x i) - (6y2 A 2x) + (2oy x ye*")
= loy — i zxy 2 + 2oyzexy.
82 MATHEMATICS FOR ENGINEERS
To find \-f-} and f j ^J x must be kept constant.
j
z = $xzy—'2x3yz-}-2oexy
then (3^- J = (5#2 X i) — (2*3 x 2y) + 20* .
3-
and
Example 17. — If z — 6 log #y — i8x5y2, find the values of [-3 — •=-)
\CLX . ^Z^ '
/ rf2^ \
and 1 , — -=- ), and state the conclusion to be drawn from the results.
\dy . dx>
To find (~j — -j-} we must first find the value of (3-), x being
regarded as a constant : then if Y be written for this expression the
value of (^ ,- j must next be determined, y being treated as a constant,
/ dzz \
and this is the value of ( , — •, ).
\dx . dyl
XT (dz\ 6xx „, 6 - - ^T
Now I T- ) = - -- i8x5X2y = --- $6x5y = Y, say.
\dxl xy y
Differentiating this expression with regard to x, y being regarded
as a constant —
or
and
\
) = —
/
. dyi \dy . dxi
Hence the order of differentiation does not affect the result.
Total Differential. — If y is a function of x, then y — f(x)
dy d fl . j.,, \
=-
i. e., dy = f'(x)dx.
dy and dx are spoken of as differentials, and f'(x) is the coefficient
of the differential dx; hence we see the reason for the term
differential coefficient.
ADDITIONAL RULES OF DIFFERENTIATION 83
If z is a function of x and y, i. e., z — f(x, y), the total differential
dz is obtained from the partial differentials dx and dy by the use
of the following rule —
fdz
dy
dy.
The reason for this is more clearly seen if we work from the
fundamental idea of rates of change, and introduce the actually
measurable quantities like Sz, Sx and Sy.
FIG. 21.
Thus —
or total change in z = change in z due to the change in
change in z due to the change in y.
The change in z due to the change in x must be measured by
the product of the change in x multiplied by the rate at which z
is changing with regard to x ; and this fact can be better illustrated
by reference to a diagram (Fig. 21).
Let P be a point (x, y, z) on a surface, and let P move to a new
position Q near to P. The change of position is made up of —
(a) A movement 8x to P' on the surface (y being kept constant) .
(b) A movement 8y to Q on the surface (x being kept constant) .
84 MATHEMATICS FOR ENGINEERS
In (a) z increases by MP'
and
i t dz \
= Sxx mean value of I -3- ).
\dx
In (b) the change in z = NQ
= Sy X mean value of ( j
\dyl
If P, P' and Q are taken extremely close to one another, the
mean or average slopes become the actual slopes and
the total change in z = 8z
-MP'+NQ = «»(*) +*(*).
YYIV
Example 18. — If Kinetic Energy = K = -- , find the change in the
energy as m changes from 49 to 49-5 and v from 1600 to 1590.
From the above rule, the change in K = 8K
s ,dK\ . s /rfK\
= 8m (dm)+8v U I
Now 8m = 49-5—49 = '5
and Sy = 1590 — 1600 = — 10.
fdK\ ,. .. d /vz \ vz
Also I j— ) (i. e., v being constant) = -j— { — X m } = — x i
\dm) v dm\2g ] 2g
(d~K.\ . , . , . d / m ,\ m
and \ WJ (m bemg constant) = dv\2> X V ) = 2 X 2V-
vm
_20 xi6oo X49
~ 64r4~ 64-4
= 19880 — 24380 = —4500 units.
Example 19. — A quantity of water Q is measured by
If rl = the probable error of D, a diameter, r.2 = the probable
error of H, a head, and R = the probable error of Q,
where ( jM) an(i (^u) are Partial derivatives.
Find an expression for R.
ADDITIONAL RULES OF DIFFERENTIATION 85
Also —
I TrCD2 /—
= — IX Vlg
Hence— R = V
R
<r
i. e., if the probable error of D is 3% and that of H is i%
that of Q = V4 (-03)«+ i(-oi)«~
= -0602, i. e., is about 6%.
Logarithmic Differentiation. — Occasionally it is necessary
to differentiate an expression which can be resolved into a number
of factors; and in such a case, to avoid repeated applications of
the rules for the differentiation of products and quotients, we may
first take logs throughout, and then differentiate, making use of
the rule for the differentiation of a function of a function. By the
judicious use of this artifice much labour can often be saved.
Example 20. — Find the value of -
Ax
Let- y = (3*~4)(4*-_
(zx-g)
then log y = log (3*— 4) +log (4*+ 7) -log (2*— 9).
Differentiating with regard to x —
dlogy _ 3 , 4_ 2
, I / i ^_\
h,,t d_l°gy dlogy dy i dy
LJUt , — r ^ - - — — . ~= —
dx dy dx y dx
i
so that
(3* -4)^(4* +7) (2* -9)
I • *?. = 3 4 2
y «** (3*-4) (4^+7) (2^—9)
^ _ (3^ -4) (4* + 7) Y
rf* ~ (2^-9)
/24*2— 66*— 189 + 24*2+ 144— 140*— 24*2— 10*4-56!
l~ (3^ -4) (4* +7) (2^ -9) I
86 MATHEMATICS FOR ENGINEERS
[As an exercise, the reader should work this according to the
following plan. Write y = -. — ^— ^r — , and then use the rule for
the differentiation of a quotient.]
It is with examples in which powers of factors occur that this
method is most useful.
T- j dy hx+2)3(x — i)
Example 21. — Find —when y = v/- --. — *-*-rs — '.
dx (2X— 5)2
Taking logs throughout —
logy = 3 log (7* + 2)+log (#—i)— 2 log (2*— 5)
Then—
djog y _ 3x7 i_ 2x2
dy ~ ~ (?x+2y(x-i) (2X-5)
— I47^r + i4^2— 31^— 10 — 28^2+2O^r+8
(*-i)(2^r— 5)
js&ar + 103
~
^i)(2X - 5)
28*2— 158^+103
y' dx~ \yx+2)(x— i)(2x— 5)
*(x-i) 28^-158^+103
Exercises 9. — On Partial Differentiation and Logarithmic
Differentiation.
1. In measuring the sides of a rectangle, the probable errors in
the sides were Y± and r2. If A = area and a and b are the sides, find
the probable error R in A.
~»(dA\*
. .
Given that- R = ^ &) +>>
the derivatives being partial.
2. If * = a-**5'*, find and
3. If 5 = /••«-#/«+log (5^-3) X««, find - and .
4. If v = (4-w)2(3 + 8w)3, find ^.
, <iy 2wy
show that - 2-
ADDITIONAL RULES OF DIFFERENTIATION 87
6. If y = 8*(i7 + -2*)«, find ^.
7. Differentiate, with respect to x,
(
8. Find the rate of discharge —j- of air from a closed reservoir
when m — -- , m, p, v and r all being variables.
CT
9. If x = r cos Q, y — r sin 0, and u is a function of both x and y,
prove that —
tdu\ t\(du\ i . Jdu
T-) = cos OU- ) — - sin
dx' \dr >o r
and
T-
\dx
du
CHAPTER IV
APPLICATIONS OF DIFFERENTIATION
HAVING developed the rules for the differentiation of the various
functions, algebraic and trigonometric, we are now in a position
to apply these rules to the solution of practical problems. By far
the most important and interesting direction in which differentiation
proves of great service is in the solution of problems concerned with
maximum and minimum values ; and with these problems we shall
now deal.
Maximum and Minimum Values. — Numerous cases present
themselves, both in engineering theory and practice, in which the
value of one quantity is to be found such that another quantity,
which depends on the first, has a maximum or minimum value
when the first has the determined value.
E. g., suppose it is desired to arrange a number of electric
cells in such a way. that the greatest possible current is obtained
from them. Knowing the voltage and internal resistance of each
cell and the external resistance through which the current is to be
passed, it is possible by simple differentiation to determine the
relation that must exist between the external resistance and the
total internal resistance in order that the maximum current flows.
Again, it might be necessary to find the least cost of a hydraulic
installation to transmit a certain horse-power. Here a number of
quantities are concerned, such as diameter of piping, price of
power, length of pipe line, etc., any one of which might be treated
as the main variable. By expressing all the conditions in terms
of this one variable and proceeding according to the plan now to
be demonstrated, the problem would become one easy of solution.
A graphic method for the solution of such problems has already
been treated very fully (see Part I, pp. 183 et seq.). This method,
though direct and perfectly general in its application, is somewhat
laborious, and unless the graphs are drawn to a large scale in
the neighbourhood of the turning points, the results obtained are
usually good approximations only. In consequence of these failings
APPLICATIONS OF DIFFERENTIATION
89
of the graphic treatment, the algebraic method is introduced, but
it should be remembered that its application is not so universal as
that of the solution by plotting.
The theory of the algebraic method can be simply explained
in the following manner :—
The slope of a curve measures the rate of change of the ordinate
with regard to the abscissa; and hence, when the slope of the
FIG, 22. — Maximum and Minimum Values.
curve is zero, the rate of change of the function is zero, and the
function must have a turning value, which must be either a
maximum or a minimum. But it has already been pointed out
that the slope of a curve is otherwise denned as the derivative or
the differential coefficient of the function ; therefore the function
has a turning value whenever its derivative is zero.
Hence, to find maximum or minimum values of a function we
must first determine the derivative of the function, and then find
MATHEMATICS FOR ENGINEERS
the value or values of the I.V. which make the derivative zero;
the actual maximum or minimum values of the function being
found by the substitution of the particular values of the I.V. in
the expression for the function.
The rule, stated in a concise form, is : To find the value of
the I.V. which makes the function a maximum or minimum, differentiate
the function, equate to zero and solve the resulting equation.
The full merit of the method will be best appreciated by the
discussion of a somewhat academic problem before proceeding to
some of a more practical nature.
Example i. — Find the values of x which give to the function
y = 2#3 + 3#2 — 36^ + 15 maximum or minimum values. Find also the
value of x at the point of inflexion of the curve.
This question may be treated from two points of view, viz. —
(a) From the graphical aspect.
We first plot the primitive curve y — 2#3+3#2— 36^+15 (see
Fig. 22), the table of values for which is : —
X
A2
Xs
2*s+3*2-36#+i5
y
-4
16
-64
— 128+48 + 144 + 15
79
-3
9
-27
- 54 + 27+108 + 15
96
-2
4
- 8
- 16 + 12+ 72 + 15
83
— I
i
- i
2+ 3+ 36 + 15
52
O
o
o
o+ o- 0 + 15
15
I
i
i
2+ 3— 36+15
-16
2
4
8
16+12— 72 + 15
— 29
3
9
27
54+27-108+15
— 12
4
16
64
128+48-144+15
47
5
25
125
250+75-180+15
1 60
This curve has two turns and two turns only, and consequently
y has two turning values, one being a maximum and one a minimum.
By successive graphic differentiation the first and second derived
curves may be drawn, these being shown on the diagram.
Now for values of x less than —3 the slope of the primitive curve
is positive, as is demonstrated by the fact that the ordinates of the
first derived curve are positive. At x — —3 the primitive curve is
horizontal and the first derived curve crosses the #-axis ; and since
dv
the ordinates of the first derived curve give the values of , , we see
dx
that when the primitive curve has a turning value, the value of
dy
•f- = o. For values of x between —3 and +2 the slope of the
primitive is negative; when x = +2 the slope is zero, and from that
APPLICATIONS OF DIFFERENTIATION 91
point the slope is positive. Thus y has turning values when x = — 3
and when x = +2; these values being a maximum at x — —3 and
a minimum at x = +2 as observed from the curve.
This investigation proves of service when we proceed to treat the
question from the algebraic aspect; in fact, for complete understanding
the two methods must be interwoven.
(b) From the algebraic point of view.
Let y = 2x3 + $x2— 36*4-15
then *£ = 6x*+6x-36
= 6(xz+x-6).
Now in order that y may have turning values we have seen that
dy
it is necessary that -^- — o.
But ^ = o if 6(#2 + *-6) = o
i. e., if 6(^+3) (x — 2) = o
i. e., if x = —3 or 2
and hence y has turning values when x = — 3 and x — +2. We do
not yet, however, know the character of these turning values, so that
our object must now be to devise a simple method enabling us to
discriminate between values of x giving maximum and minimum
values to y.
An obvious, but slow, method is as follows : Let us take a value
of x slightly less than —3, say —3-1; then the calculated value of y
is 95 "85. Next, taking a value of x rather bigger than —3, say —2-9,
the value of y is found to be 95-85. Therefore, as x increases from
— 3-1 to —3 and thence to —2-9, y has the values 95'85, 96, and
95-85 respectively. Thus the value of y must , be a maximum at
x = — 3, since its values on either side are both less than its value
when x = — 3. In like manner it can be shown that when x — -\-2,
y has a minimum value.
The arithmetical work necessary in this method can, however,
be dispensed with by the use of a more mathematical process,
now to be described.
Referring to the first derived curve, the equation of which is
y — 6#2+ 6#— 36, we note that as x increases from —4 to —3
the ordinate of the derived curve decreases from 36 to o; from
x = — 3 to x = — .5 the ordinate is negative but increasing
numerically, i. e., in the neighbourhood of x = — 3 the slope of
the second derived curve, which is the slope curve of the first
derived curve, is negative (for the ordinate decreases as the
abscissa or the I.V. increases). But the slope of the first derived
92 MATHEMATICS FOR ENGINEERS
curve, and thus the ordinate of the second derived curve, must
d2v
be expressed by -y-^, so that we conclude that in the neighbourhood
dsC
of a maximum value of the original function the second derivative
of it is a negative quantity.
In the same way we see that in the neighbourhood of a
minimum value of the function, its second derivative is a positive
quantity. Hence a more direct method of discrimination between
the turning values presents itself : Having found the values of the
I.V. causing turning values of the original function, substitute these
values in turn in the expression for the second derivative of the function ;
if the result is a negative, then the particular value of the I.V. considered
is that giving a maximum value of the function and vice-versa.
This rule may be expressed in the following brief fashion : —
Let y = f(x) and let the values of x that make -j~(x) orf'(x) = o
be #! and xz.
d2y
Find the value of -~^ or f"(x), as it may be written, and in
this expression substitute in turn the values x^ and xz in place of
x: the values thus obtained are those of f"(x-^} and /"(#2) respec-
tively. Then if f"(x^), say, is negative, y has a maximum value
when x — x^; and if f"(xj) is positive, y has a minimum value
when x = xv
Applying to our present example : —
y =
/
When x — —3 the value of -~2 is I2( — 3) +6, i. e., /"(— 3) = —30;
and since /"( — 3) is a negative quantity, y is a maximum when
x == —3.
Similarly, /"( + 2) = 12(2) +6 = +30
and hence y is a minimum when x — +2.
Referring to the second derived curve, i. e., the curve y = I2X+6,
we note that its ordinate is negative for all values of x less than
— •5 and positive for all values of x greater than —-5, the curve
crossing the axis of x when x = —-5. This indicates that when
x = — -5 the first divided curve has a turning value ; but the first
APPLICATIONS OF DIFFERENTIATION 93
derived curve is the curve of the gradients of the primitive curve,
and hence when x — — 5 the gradient of the primitive must have
a turning value, which may be either a maximum or a minimum.
In other words, if we had placed a straight edge to be tangential
in all positions to the primitive curve, it would rotate in a right-
handed direction until x = — -5 was reached, after which the
rotation would be in the reverse direction. A point on the curve
at which the gradient ceases to rotate in the one direction and
commences to rotate in the opposite direction is called a -point of
inflexion of the curve. Thus points of inflexion or contra-flexure
u d*y
occur when -=-4 = o.
dx2
A useful illustration of the necessity for determining points of
contra-flexure is furnished by cases of fixed beams. We have
of Confraf lexure.
[—• filll— 1 /^ ^\ I— -21 ll— !
FIG. 23.
already seen that the bending moment at any section is propor-
d2v
tional to the value of -j-z there; hence there must be points of
CLX*
contra-flexure when the bending moment is zero.
Example 2. — Find the positions of the points of contra-flexure of
a beam fixed at its ends and uniformly loaded with w units per foot;
the deflected form having the equation —
i fwlx3 wl2xz_wx*\
= El\ 12 24 24 ')'
We may regard this question' from either the graphic aspect or the
physical. According to the former we see that it is necessary to
determine the points of inflexion, and therefore to find values of x
^ ^d*y .
for which , 2 is zero.
Reasoning from the physical basis we arrive at the same result,
by way of the following argument : the bending moment, which is
dzy
expressed by EI-— , changes sign, as is indicated by the change in
OLX
the curvature of the beam (see Fig. 23), and therefore at two points
the bending moment must be zero, since the variation in it is uniform
94 MATHEMATICS FOR ENGINEERS
and continuous; but the bending moment is zero when -~2 is zero,
v
since M = El j-^.
dx2
AT w fix3 I2x2 x*\
Now y = =pp ( ----------- V
El \i2 24 24/
dy w T( I ,\ / /2 \ 4^
hence ^ = -ey ( — X3*2 ) — ( — X2# )— *-
a* El LA 1 2 / \24 / 24
w __
12 ~~ 6
d2y w r / 1 \ i lz \ 3^2-i
and -,- -2 = ,-,f ( xzx )— ( XII — V
^2 El L\4 / \I2 / 6 J
— (**— l* _^2\
. El Va i2~ 2"]'
/72V ./v 72 «;2\
Now the bending moment M = EI.^=a;[-— -- - )
«*" \ 2 12 2 /
/^ /2 AT2
and M = o if ------ , i. e., 6lx—l2—6xz = o,
2 12 2
i. e., if 6xz— 6lx+lz = o
6l±
or
12
= -789? or -2 1 iL
Hence the points of inflexion occur at points distant -211 of the
length from the ends.
Example 3. — A line, 5 ins. long, is to be divided into two parts
such that the square of the length of one part together with four
times the cube of the length of the other is a minimum. Find the
position of the point of section.
Let x ins. = the length of one part, then 5— x = length of the
other part.
Then (5— #)2+4#3 is to be a minimum. *
Let y = (5-*)2+4*3
Then
Hence ~- — o if x — ^ or — i (the latter root implying external
dx o
cutting) .
APPLICATIONS OF DIFFERENTIATION 95
To test for the nature of the turning value —
— 10
dx
d*y
and g£ = 24*4-2.
When x = J
6
-5-^ = ( -2— —5 J_|_2 = a positive quantity.
Therefore y is a minimum when x = £ and the required point of
section is % in. from one end.
Example 4. — If 5 — detrimental surface of an aeroplane
S = area of planes
K = lifting efficiency
KS
then/, the " fineness," is obtained from the formula /2 = — ~-.
Also the thrust required for sustentation = C (••I'acX where C is
a constant and i is the angle of incidence of the plane (expressed in
radians).
Taking S = 255 and K = -4, find the angle of incidence for the
case in which the least thrust is required.
p = J£S .4X25 =
J -o8s -08 5'
The thrust T = C (*4-js:.) and since i is the only variable in this
\ J *%/
expression, we must differentiate with regard to it.
Thus -
dT .,
and = 0 if 1- =
i. e., if i2 = 1 = — .
/2 125
Thus i = -0895
or the thrust required is either a maximum or minimum when the
angle of incidence is -0895 radian.
To test whether this turning value is a maximum or a minimum,
let us find the second derivative —
^-
dT=
^T _ / 2_
&-« - u V +/2z3/'
^2-p
When i — -0895, -^ must be positive, and hence T has its minimum
value when i = -0895.
96 MATHEMATICS FOR ENGINEERS
Example 5. — Find the dimensions of the greatest cylinder that
can be inscribed in a right circular cone of height 6 ins. and base
10 ins. diameter.
FIG. 24.
Assume that the radius of the base of the cylinder = x ins. (Fig. 24)
and the height of the cylinder — y ins.
Then the volume — V — •nx^y.
We must, then, obtain an expression for y in terms of x before
differentiating with regard to x.
From the figure, by similar triangles, taking the triangles ADC
and EFC —
6 _ y
or
Hence
V
5
2 6 . . 67T , g 3,
— it* X-(5 • ~~ ~c '*•* , •
and
Thus
dV
d\'
5
= o if x(io — 3^) = o
i. e.t
or
if x = o (giving the cylinder of zero
if 10 — 3*, i. e., x — $\ ins.
volume)
Then
y = (5-3*) = 2 ins-
and the volume of the greatest cylinder = wX ( — ) X2 = 69-8 cu. ins.
Example 6. — The total running cost in pounds sterling per hour
of a certain ship being given by —
v3
C =4.5+—
^2100
where v = speed in knots, find for what speed the total cost for a
journey is a minimum.
The total cost for the journey depends on —
(a) The cost per hour; and
(6) The number of hours taken over the journey.
APPLICATIONS OF DIFFERENTIATION 97
Item (b) depends inversely on the speed, so that if the journey
j t
K
were 2000 nautical miles the time taken would be hours; or.
in general, the number of hours =
Then the total cost for a journey of K nautical miles
K / v3 \
= Q = xU-5+-
V V* ' ' 2IOO/
i , UM
5V -\ )•
2IOO/
Differentiating with regard to the variable v
dCt
Then
1050
.,
* a
v • (s . . IX
"4-5
- ' *J
-
1050 v
or v3 = 4-5 X 1050 = 4725
hence *v = 16-78 knots.
Example 7. — A water main is supplied by water under a head of
60 ft. The loss of head due to pipe friction, for a given length, is
proportional to the velocity squared. Find the head lost in friction
when the horse-power transmitted by the main is a maximum.
If v = velocity of flow, then —
Head lost = Kv2, where K is some constant,
i. e., the effective head = 60 — Kv2 = He.
TT -r, , Quantity (in Ibs. per min.) x effective head (in feet)
H.P. transmitted = J v * < - v '
33000
_ area (in sq. ft.) x velocity (ft. per min.) x 62-4 x Hg
33000
= CvHe, where C is some constant
= Gv{6o-Kv*)
= C(6ov-Kv3)
Then / (H.P.) = C(6o-3Kz;2)
Ctl)
= C(6o — i8o + 3He)
or T— (H.P.) = o when 3HC = 120
i. e., He = 40.
In general, then, the maximum horse-power is transmitted when
the head lost is one-third of the head supplied, i. e., the maximum
2
efficiency is - or 66-7%.
98 MATHEMATICS FOR ENGINEERS
Example 8. — The stiffness of a beam is proportional to the breadth
and the cube of the depth of the section. Find the dimensions of
the stiffest beam that can be cut from a cylindrical log 4 ins. in
diameter.
From hypothesis
or
S = Kbd3.
Both breadth and depth will vary, but they depend on each
other; and from Fig. 25 we see that b2 = i6 — d2. Hence we can
substitute for b its value in terms of d and then differentiate with
regard to d\ according!}? —
S =
As it stands this would be a rather cumbersome
expression to differentiate, and we therefore employ
a method which is often of great assistance. Since
we are dealing with positive quantities throughout,
S2 will be a maximum when S is a maximum,* and
hence we square both sides before differentiating.
Thus— S2 = K2d6(i6-d2) =
<fS2
and -, ~
FIG. 25.
'>-8d'>) = 8d5(i2-d2)
Hence — ,-r = o if d5 — o, i. e., d = o (giving zero stiffness)
Ct.(t
or if —
i. e.,
Hence
d2 = 12
d = 3-464 ins.
b — V 16—12 = 2 ins.
* If we were dealing with negative quantities it would be incorrect
to say that the quantity itself had a maximum value when its square
was a maximum, for suppose the values of the quantity y in the
neighbourhood of its maximum value were —13, — 12, — u, — 10,
— ii, —12, etc., corresponding values of y2 would be +169, +144,
+ 121, +100, +121, +144, so that if y = — 10 (its maximum value)
when x = 4, say, then y2 — TOO when x = 4, and therefore a minimum
value of y2 occurs when x — 4, and not a maximum.
Example g. — Find the shape of the rectangular channel of given
sectional area A which will permit the greatest flow of water ; being given
that Q = Av, v — c Vmi, m = hydraulic mean depth = — . . f^rea.
wetted perimeter
and i is the hydraulic gradient ; Q being the quantity flowing.
APPLICATIONS OF DIFFERENTIATION 99
Let the breadth of the section be b and the depth d; then, by
hypothesis —
bd = A. whence b = -j.
a
w = — . . , — -. — — = ,— — j and therefore v = c Vi\/ ;
wetted perimeter o+2a
= cVAl .
Hence — Q = Av = Ac VAi . ~ /, ,
= K . -^Jsas where K =
Vb+2d
Q will be a maximum when Q2 is a maximum, hence we shall find
the value of b for which Q2 is a maximum.
b+2d~ '6+?A'
Also Q2 is a maximum when the denominator of this fraction is
a minimum.
Let this denominator be denoted by D —
dD d I, , 2A\ 2A
then -jjj- = ^1°+ r / = I~~^F
rfD ., 2A . /— »—
and -n- = o if i = T-», «. e., if o = V2A.
Now d = = -= = \
O \/2A 2
/. the dimensions would be —
/A
= -
depth = V - and breadth = \/2A.
Example 10. — For a certain steam engine the expression for W,
the brake energy per cu. ft. of steam, was found in terms of r, the
ratio of expansion, as follows —
/i+log r\
( r— S-)-27
I2o
\V =
- _ - _ - _
00833 ,
JJ.J-. 000903
Find the value of r that makes W a maximum.
Before proceeding to differentiate, we can put the expression in a
somewhat simpler form.
Thus- W = "O
•00833 +
and W is a quotient = - where u = i2o(i+log r}—2jr
ioo MATHEMATICS FOR ENGINEERS
du 1 20
so that -y— = ---- 27
dr r
and v — -00833 + -000903?
dv
so that -f- — -000903.
du dv
J\*T VJ -- MJ~
TT dW dr dv
Hence -y = — , —
dr vz
(•00833 + •000903*') ( — — 27^ — [iao(i+log r)— 27/1-000903
(-00833 +-000903*-)2
Now - = o if the numerator of the right-hand side = o
i . e., if ( -—22 — L°\_ (27 x -00833) + (120 x -000903) — (27 x -0009037)
— (120 x -000903) — (120 x -000903 log r) + (27 x -ooogo^r) = o
i.e., if. —-225 — -1084 log r = o.
This equation must be solved by plotting, the intersection of the
curves y± = -1084 log r and yz = --- 225 being found; the value of r
here being 2-93.
Hence — r = 2-93.
Example n. — The value of a secondary electric current was given
by the formula —
_- _
y = - e L+M— e L-M
where L = inductance of primary circuit
R = resistance of primary circuit
M = coefficient of mutual inductance
I = steady current.
Find for what value of /, y has a maximum value.
T / iu m
y = -(e~L+M— g~L— M
dv R — ?L R -
and -£ = o if ^^r.e L-M — __
rf/ L— M L+M
Transposing the factors —
L— M
BI(L-M-L-M) T _
L+M
-TM, ~
e L-M. =
or eLM.
L— M
APPLICATIONS OF DIFFERENTIATION 101
In order to find an expression for t, this equation must be changed
to a log form, thus —
/L+M\ _ 2MRf
°g \L— M/ ~ L8-M2
L2-M2 , /
* = - log
If three variables are concerned, say x, y and z, the relation
between them being expressed by the equation z=f(x, y), then
in order to find the values of x and y for turning values of z, it
is necessary to determine where the plane tangential to the surface
is horizontal.
The algebraic problem is to find the values of x and y that
satisfy simultaneously the equations (-T-) = o and f-^-J = o, these
\(4'Z / \dZ /
derivatives being partial.
Example 12. — The electric time constant of a cylindrical coil of
wire (i. e., the time in which the current through the coil falls from
its full value to a value equal to -632 of this) can be expressed
approximately by K = — . y — where z is the axial length of the
coil, y is the difference between the external and internal radii and x
is the mean radius ; a, b and c representing constants. If the volume
of the coil is fixed, find the values of x and y which make the time
constant as great as possible.
The volume V of the coil = cross section x length
V
*. e., V = 2Tc#xy Xz and z = -
•2-nxy
K = m \ — r<--- I and is a maximum
(ax+by+cz)
ax+by+cz a , b , c '.
when - or — -\ — is a minimum.
xyz vz xz xv
f , a. . b , c
Let p = — --- —
z xz x
- — - --- —
xyz yz xz xy
a. . b , c
— --- —
yz xz xy
. c
+
yV xV x~y
__ zivcya mxyb c
'' ~ +"
Now (^ (i. e., with y constant) = 2yU+(- X -
2TOI C
102 MATHEMATICS FOR ENGINEERS
c- -11 (dp\ 27T& C
Similarly \f) = -~- 2.
\dy)x V xyz
Both \j) an<^ vft mus^ ke equated to zero,
so that —^jr- — —
V x2y
i. e., x*y = — (i)
, 2TI& C
V ' xy2
i. 6. xyz = .... (2)
2TT&
To solve for x and y —
cV
From (2) — x = — j-~2.
Substituting in (i) —
cV
cVa
whence y3 = — r,
27T02
or
__2TU62
also
Exercises 10. — On Maximum and Minimum Values.
1. If M = 15*— -oix*, find the value of x that makes M a maximum.
2. Find the value of x that makes M a maximum if M = 3-42* — -ix2.
3. M is a bending moment and x is & length ; find x in terms of /
so that M shall be a maximum, and find also the maximum value of M.
M = <
4. As for No. 3, but taking —
5. The work done by a series motor in time t is given by —
,, wx ,,
M = — (l—
2 V
R
where e = back E.M.F.
E = supply pressure
R = resistance of armature.
a
The electrical efficiency is ^. Find the efficiency when the motor
so runs that the greatest rate of doing useful work is reached.
In Nos. 6 to 8 find values of x which give turning values to y,
stating the nature of these turning values.
APPLICATIONS OF DIFFERENTIATION
6. y = 4*2+i8*-4i. 7. y = 5^-
8. y = x3 + 6x2— 15^+51 (find also the value of x at the point of
inflexion).
9. Sixteen electric cells, each of internal resistance i ohm and
giving each I volt, are connected up in mixed circuit through a
resistance of 4 ohms. Find the arrangement for the greatest current
[say — rows with x cells in each row].
x
10. If 40 sq. ft. of sheet metal are to be used in the construction
of an open tank with square base, find the dimensions so that the
capacity of the tank is a maximum.
fj A
11. Given that W = 4C2+',7, find a value of C that gives a turning
value of W, and state the nature of this turning value.
// _ x\
12. M (a bending moment) = W v . ; (x+y] — Wy. For what value
of x is M a maximum ? {W, / and y are constants.}
13. The cost C (in pounds sterling per mile) of an electric cable
can be expressed by —
C - ^+636*
x
where x is the cross section in sq. ins.
Find the cross section for which the cost is the minimum, and
find also the minimum cost.
14. A window has the form of a rectangle together with a semi-
circle on one of its sides as diameter, and the perimeter is 30 ft. Find
the dimensions so that the greatest amount of light may be admitted.
15. C, the cost per hour of a ship, in pounds, is given by —
c3
C = 3-2 + -
2200
where s = speed in knots.
Find the value of s which makes the cost of a journey of 3000
nautical miles a minimum.
At speed 10% greater and less than this compare the total cost
with its minimum value.
16. An isolated load W rolls over a suspension bridge stiffened
with pin-jointed girders. When the load is at A, distant x from the
\V#
centre, the bending moment at this section — MA = —» (I2— 4#2). For
what value of x is MA a maximum ?
17. A riveted steel tank of circular section open at the top has to
be constructed to contain 5000 gals, of water. Find the dimensions
so that the least possible amount of steel plate is required.
18. A canister having a square base is cut out of 128 sq. ins. of
tin, the depth of the lid being i in. Find the dimensions in order
that the contents of the canister may be as large as possible.
19. The stiffness of a beam of rectangular section is proportional
to the breadth and the cube of the depth. Find the ratio of the
sides of the stiffest beam of rectangular section with a given perimeter.
104 MATHEMATICS FOR ENGINEERS
20. A load uniformly distributed over a length r rolls across a
beam of length I, and the bending moment M due to this loading at
a point is given by —
-„. wry f, r} wxz
M=-j* {/-?+*--,- — .
For what value of x is M a maximum ?
21. Find the value of V (a velocity) that makes R (a resistance)
a maximum when —
= yi 3(V-i2)
54 V+I2 '
22. If L = Vrz— xz— *(rz— x*), find the value of x that makes L a
yv
maximum.
23. A jet of water, moving with velocity v, impinges on a plate
moving in the direction of the jet with velocity u. The efficiency
TJ = --—3 — - • Find values of u for maximum and minimum efficiency,
and find also the maximum efficiency. .
^2.tt(l} _ /vC\
24. If v — — *-= — ', find the value of u for maximum value of i?.
vz
25. Given that O = K//T! (cos 6— sin 6), find values of 6 between
o° and 360° that make Q a maximum, treating K, p. and Tx as
constants.
26. A cylinder of a petrol engine is of diameter d and length /.
„. , , d area of exposed surface
Find the value of the ratio -, which makes
/ capacity
a minimum. The volume must be treated as a constant.
27. If the exposed surface of a petrol engine cylinder is given by —
S — 27W2+2:rr/+-2y2, I being the length and r the radius,
find the value of the ratio - that makes the ratio exPosed surface
r capacity
a minimum. The volume must be treated as a constant.
»
28. Given that ? = — K2 - , find values of K for turning
values of y.
z#R2/ 1 \ / /3\
29. IfM = ^|^ -sin20)--934wR2^cos0- /s/|j, for what values
of 0 is M a maximum ? [M is the bending moment at a section of a
circular arched rib loaded with a uniform load w per foot of span, and
R is the radius of the arch.]
30. An open channel with side slopes at 45° is to have a cross
section of 120 sq. ft. Determine the dimensions for the best section
(i. e., the section having the smallest perimeter for a given area).
_
31. If M = -^TTZ - ni\ * find the value of x which makes M, a
O ( I — .- t )
bending moment, a maximum. The final equation should be solved
by plotting, a value being assumed for /.
APPLICATIONS OF DIFFERENTIATION 105
32. In connection with retaining walls the following equation
occurs —
p _ ph2 _ i — M tan 6
72 i— M*+2/* cot 6.
Find an expression, giving the value of 6 (in terms of tan 6), that
makes P a maximum. {M, p and h are constants.}
33. Assuming that the H.P. of an engine can be expressed by the
relation —
H = C(fnl3-Kpn3l*)
where C is a constant, / = stroke, p = pressure in piston rod due to
the pressure on the piston, p — average density of the material of the
engine, K = constant depending upon the mode of distribution of
the mass of the engine parts, n = R.P.M., and / = safe stress in the
material, find an expression for I giving the maximum H.P. for
engines of different sizes.
34. Find the turning point of the probability curve —
T *»
and also the points of inflexion.
35. In a two-stage compressor, neglecting clearances, if Px and Vx
are the initial pressure and volume of the L.P. cylinder, P2 the pressure
in the intercooler, and P3 the discharge pressure of the H.P. cylinder,
the total work for the two cylinders is given by —
For what value of P2 is W a minimum, P^ V^, P3 and n being
regarded as constants ?
36. Find the height A of a Warren girder to give the maximum
stiffness, the stiffness being given by the expression —
tln , Id { W\
2EA r • 4
d being the length of one bay and I the span, whilst fc, ft and E are
constants for the material.
37. The efficiency of a reaction wheel may be expressed by —
2(«— i)
For what value of n has •>> its maximum value ?
38. The weight W of steam passing through an orifice, from
pressure P! to pressure Pg, is given by —
n+Pj"
If n — 1-135, nn(i the value of ^ for which W is a maximum.
*i
39. Find the height of the greatest cylinder that can be inscribed
in the frustum of a paraboloid of revolution cut off by a plane
perpendicular to the axis and distant 6 units from the origin. The
paraboloid is generated by the revolution about the axis of x of the
parabola yz = yc.
106 MATHEMATICS FOR ENGINEERS
f 1% _!_ y\ ~\
40. If M = \V j# — - — 1*-\ where y and I are constants, find the
value of x that makes M a maximum.
41. If T, / and T/ are the tight, slack and centrifugal tensions
respectively in a belt passing round a pulley, and v = speed of the
belt in feet per sec., then —
H.P. transmitted - ^1=3.
550
TJOV^
Being given that T/ = — , the maximum permissible tension in
the belt — Tm = T+T/, ^ = coefficient of friction between belt and
T
pulley, 0 = angle of lap of belt in radians, and . — eve, find the value
%
of T/ in terms of 1m so that the maximum H.P. is transmitted.
42. If y — 3X*-\-2x3 — j8x2—24OX-\-54, find the values of x which
give turning values to y, stating the nature of these turning values ;
and find also the values of x at the points of inflexion.
43. The radial stress in a rotating disc
in which expression x is the only variable.
Find the value of x which gives to px its maximum value, and
state this value of px.
44. A pipe of length / and diameter D has at one end a nozzle of
diameter d through which water is discharged from a reservoir, the
level of the water in which is maintained at a constant head h above
the centre of the nozzle. Find the diameter of the nozzle so that
the kinetic energy of the jet may be a maximum ; the kinetic energy
being expressed by —
7T V_2gD'A \t
[Hint. — If K = kinetic energy, writ
_ P7T
=
and find the value of d for the maximum value of K*.]
45. Prove that the cuboid of greatest volume which can be inscribed
in a sphere of radius a is a cube of side
46. The velocity of the piston of a reciprocating engine can be
expressed by —
/sin 20 .. Q\
2nnr\ — -- hsm 0 )
\ •zm I
where 0 is the inclination of the crank to the line of stroke.
T ,. connecting-rod length 0 _ , .
If m = - TxrS ---- 1 - — = 8, find the values of 0 between
length of crank
o° and 360° that make the velocity a maximum.
APPLICATIONS OF DIFFERENTIATION
107
Calculation of Small Corrections. — Differentiation finds
another application in the calculation of small corrections.
Thus an experiment might' be carried out, certain readings
being taken, and results deduced from these readings; then if
there is a possibility of some slight error in the readings and it
is required to find the consequent error in the calculated result,
we may proceed to find that error in the manner now to be
explained.
Suppose we have two quantities A and B connected with one
another by a formula A = KB ; then if the value of B is slightly
inaccurate the error in A will depend on this error in B, and also
on the rate at which A changes with regard to B. E. g., if A
changes three times as fast as B and the error in B is •!%, then
the consequent error in A must be 3X-I or -3%.
We might also look upon this question
from a different point of view. Suppose
that a reading, instead of being x, as it
should have been, was slightly larger,
say x-}-8x, i. e., the measured value of x
would be represented by OB and not
OA (Fig. 26), then the error is Sx or
8x
— XIOO%.
X
This error causes an error
in the value of y, so that the calculated
value of y is BQ and not AP, i. e., the
error is Sy.
To compare these errors we proceed as follows : -- — the slope
of the chord PQ, and if Sx is very small (as it should be, for
otherwise the experiment would be repeated), then this would also
be the slope of the tangent at both P and Q, or, approximately —
Sy dy
• 8x dx
^. e.,
dx
or, error in y = rate at which y changes with regard to x
X error in x.
Example 13. — In the measurement of the diameter of a shaft, of
which the actual diameter was 4 ins., an error of 2% was made;
what was the consequent error in the weight ?
io8 MATHEMATICS FOR ENGINEERS
Here — W = ~d2lp, where p is the density
4
- Kd2, where K = %.
4
Now the error in the diameter = Sd = --'- X4 = -08 in.
too
also - - - 2Kd
CLLJ\J i i - 7 i - £• J.Vli-
d.d d.d
or the percentage error = ^ x 100 = - ^9g^- X 100
Example 14. — If some torsion experiments are being made on
shafts varying in diameter from i in. to 5 ins. ; then, allowing a maximum
error of -5% in the measurement of the diameters, what is the range
of the errors in the stress ? Given that T = f-fd3.
The stress / = — x -^
TC a3
df i6T
hence 13- —
Now the error in the diameter — 8.d is -5%
i. g., 8d = — 5 X rf.
100
Hence the "error in / = /-. X 8d = — 4-7r X —
d.d -jzd* 100
i. e.t the percentage error in / —
8f 48T -5d
= loo X Jf- = loo X — —r, X ~ X
/ IT a4 zoo
= -3-
Thus the smallest error = '03 x smallest stress \
and the largest error = -03 x largest stress /'
If the error in the measurement of the diameter is on the high
side, then the stress, as calculated, will be too low.
Expansion of Functions in Series. Theorems of Taylor
andMaclaurin. — Many of the simpler functions, suchaslog« (i-f-#),
sin x, cos x, etc., can be expressed as the sums of series. These
functions can be expressed in terms of these series by the use of
a theorem known as Maclaurin's.
APPLICATIONS OF DIFFERENTIATION 109
Let f(x) stand for the function of x considered, and let
f(x) — a+bx+cx^+dx3-^ . . ., to be true for all values of x,
i.e., /(o) = «
We assume that the differentiation of the right-hand side term
by term gives the derivative of /(#).
Differentiate both sides with regard to x.
Then— ^or/'(*) = b+2cx+idx*+ . . .
This must be true for all values of x ; thus, let x = o
then — f'(x) when x = o dr/'(o) = b
/'(o) implying that f'(x) or •£*-*• is first found and then the
flwv
value o substituted for x throughout.
d*f(x)
Differentiating again, -4V orf"(x) = 2c-}-6dx-{- . . .
and /"(°) = zc
/"(o)
t. e., c=J—±J.
2
Similarly, /'"(*) = 6^+ terms containing x and higher
* powers of x,
whence /'"(°) = 6W or 1.2.3.^.
or
1.2.3 |JL
Accordingly we may write the expansions —
This is Maclaurin's Theorem. By a similar investigation we
might obtain Taylor's Theorem, which may be regarded as a more
general expression of the foregoing.
Taylor's Theorem. In this the expansion is of f(x-\-h] and not
/(*); thus—
\A
or, as it is sometimes written, to give an expansion for/(*) —
no MATHEMATICS FOR ENGINEERS
If in either of these two expansions we make h = o, then
Maclaurin's series results.
We may now utilise these theorems to obtain series of great
importance.
Example 15. — To find a series for cos x.
Let f(x] = cos x
then /(o) — cos o = i.
A i rn \ • d COS X
Also / (x), i. e., — -, -- = —sin x
ctx
so that /'(°) = —sin o — o.
Again /"(•*) = ^~(~s^n x) ~ —cos x
so that /"(°) = —cos o = — i
and /"'(#) = -3- (—cos x) — sin x
so that /'"(°) = sin o = o.
Now f(x] = /(o
Therefore cos x — i --- 1
12. [4
Example 16. — To find a series for loge
Let
then
Now
so that
so that
and
so that
Hence
/(*)
/(°)
/'(o)
/"(o)
f"'lv\
J (*/
/'"(o)
loge (i+#)
= loge(i+#)
= log 1=0.
rflog(l+#) I
~ i
i
d/ -i \ 2
^\(l+^)2/ (l+^)3
2
= - — 2.
^r2 Ar3
(Compare with the series found by an entirely different method in
Part I, p. 470.)
Example 17. — Prove that &>* = cos #+_;' sin ^, where /= V— I.
This equation is of great importance, since it links up the exponential
and the trigonometric functions.
APPLICATIONS OF DIFFERENTIATION in
To find a series for sin x.
Let . f(x) = sin x then /(o) = sin 0 = 0
f'(x] = cos x /'(o) = cos o — i
f"(x) = —sin x /"(°) — —sin o = — o
f"'(x) = —cos x f "'(o) = —cos o = — i.
Hence sin # = x 1 . . .
LI LL
and jsinx = i(x |--j . . .)
v Li U
To find a series for eix.
Let f(x) = eix /(o) = e° = i
f"(x) = jzeix /"(o) = jze° = jz = — i
-7 2 A • 2 ^ ^X^
Hence eix — i+jx+- \-J- — . . .
11 11
Now cos x+j sin # (the series for cos x having been found in
Example 15).
Ivllj|+ " { Fory: : -_\^ \
= eix. ( i*= +i, etc. J
Use might be made of Taylor's Theorem to determine a more
correct solution to an equation when an approximate solution is
known ; for, taking the first two terms of the expansion only —
/(*+*)=/(*)+*/'(*)
or interchanging x and h, as a matter of convenience, then —
/(*+*)=/(*)+¥'(*)•
If A is small compared with x, the assumption that two terms
of the series may be taken to represent the expansion is very
nearly true.
Suppose that a rough approximation for the root has been
found (by trial and error) ; denote this by x. Let the true solution
be x-\-h; then by substitution in the above equation the value of
h can be found, and thence that of x-}-h.
H2 MATHEMATICS FOR ENGINEERS
As an illustration, consider the following case : A rough test
gives 2-4 as a solution of the equation x*— i -5^+3 -7% = 21-554.
It is required to find a solution more correct.
Here x = 2-4 and f(x) = #4— 1-5#3+37#— 21-554
so that 7(2-4) = 33-17— 20-73+8-88— 21-554 = -'234-
If the correct value of h is found, thenf(x-\-h) must = o.
Hence— f(x+h) =f(x)+hf'(x)
i. £., o = — 234+A/'(2-4).
[Now— /(*) = *4—i-5*3+37*— 21-554
so that 7' (2-4) = 55-30-25-92+3-7 = 33-o8.]
Hence o = — 234+^x33-08)
or h= — -^o = -0071.
33-o8
Hence a more correct approximation is 2-4+ -0071
*. e.t x = 2-407 is the solution of the equation.
This method may thus be usefully employed in lieu of the
graphic method when extremely accurate results are desired.
The following example illustrates the process of interpolation
necessary in many cases where the tables of values supplied are
not sufficiently detailed for the purpose in hand; and in view of
the importance of the method, every step in the argument should
be thoroughly understood.
Example 18.' — It is desired to use some steam tables giving the
pressures for each 10° difference of temperature, to obtain the accurate
d-b
value of •£. when t = 132° C. The figures in the line commencing
with i = 130° C. (the nearest to 132°) are as follows : —
1
dp
• : V
d*p
ip
d*P
dp"
d'p
dt*
130
2025-7I7 60-5995
1-47051
•026392
•0002738
Calculate, very exactly, the value of -^ when t = 132° C.
Taylor's theorem may here be usefully employed, using the form —
APPLICATIONS OF DIFFERENTIATION 113
Let f(x) = -£- when t — 130
andf(x + h) = - when t = 132, so that h = 2
Then /'(*) = = and /"(*) = , etc., / having the
value 130.
Thus the expansion may be re-written as —
_fdp
and substituting the values from the table —
= 60. 5995 + (2 x.i-4705i) + (2 X -026392) + (4 x -0002738)
'133 V3
= 63-59367-
Exercises 11. — On the Calculation of Small Corrections and Expansion in
Series.
1. If R = R0(i+a/+6*2) when R0 (the resistance of a conductor at
o° C.) is 1-6, a (the temperature-resistance coefficient of the material)
= -00388 and b = -000000587, find the error in R (the resistance at
temperature t° C.) if t is measured as 101 instead of 100.
2. The quantity Q of water flowing over a notch is given by
o -
Q = — X-64X V^g.H*, where H is the head at the notch. What
is the percentage error in Q caused by measuring H as -198 instead
Of -2 ?
3. If y = 4*1'76, y = 17-3 when x = 2-3. What will be the change
in y consequent on a change of x to 2-302 ?
4. A rough approximation gives x = —2-44 as a solution of the
2j
equation 10 3 = 16+4^— xz. Find a more correct root.
5. Determine the value of x to satisfy the equation x1'5— 3 sin x =• 3,
having given that it is in the neighbourhood of 2-67.
6. The height A of a Porter governor is expressed by —
w ' n*
where n is the number of revolutions per minute. If W = 100, w = 2
and / = 10, find the change in the height due to a change in the
speed from 200 to 197 r.p.m.
7. In calculating the co-ordinates of a station in a survey it was
thought that there was a possibility of an error of 3 minutes (*. e., i£
either way) in the reading of the bearing. If the bearing of a line
was read as 7° 12' and the length of the line was 2 chains 74 links,
I
n4 MATHEMATICS FOR ENGINEERS
find the possible errors in the co-ordinates of the distant end of the
line. [Co-ordinates are length X cos (bearing) and length x sine (bearing) .]
8. Find by the methods of this chapter a series for ax.
9. Using the figures given in Example 18, p. 112, calculate very
exactly the pressure p at 133° C.
10. The equation d3+-6$d— -5 = o occurred when finding the sag
of a cable. A rough plotting gives the solution to be in the neigh-
bourhood of -5 : find a more exact root.
CHAPTER V
INTEGRATION
HAVING discussed the section of the Calculus which treats of
differentiation, we can now proceed to the study of the process
of integration, this having a far more extensive application, and
being, without doubt, far more difficult to comprehend.
As with the differentiation, it is impossible fully to appreciate
this branch of the subject unless much careful thought is given
to the fundamental principles; and accordingly the introduction
to the Integral Calculus is here treated at great length, but in a
manner which, it is hoped, will commend itself.
Meaning of Integration. — The terms integer and integral
convey the idea of totality; an integer being, as we know, a
whole number, and thus the sum of its constituent parts or
fractions. The process of integration in the same way implies a
summation or a totalling, whereas that of differentiation is the
determination of rates of change or the comparison of small
differences. Differentiation suggests subtraction or differencing,
whilst integration suggests addition ; differentiation deals with rates
of change, integration with the results of the total change ; differen-
tiation involves the determination of slopes of curves, and integra-
tion the determination of areas of figures. Integration is, in fact,
the converse to differentiation, and being therefore a converse
operation is essentially more difficult to perform. [As instances
of this statement contrast the squaring a quantity with the
extraction of a square root, or the removal of brackets with
factorisation.]
A converse operation is rather more vague as concerns the
results than a direct; for when performing a direct operation one
result only is obtainable, but the results of a converse operation
may be many, as we shall find, for example, when dealing with
indefinite integrals.
To illustrate the connection between differentiation and integra-
tion, consider the familiar case of velocity and acceleration. Suppose
values of v and t are given, as in the table : —
"5
n6
MATHEMATICS FOR ENGINEERS
Then
t
•I
•15
•20
•25 -30
V
28-4
297
30-5
33'4 36-5
8v
i'3
•8
2-9
3'i
8t
•05
•05
•05
•05
Bv
a = w
26
16
58
62
The accelerations are here found by comparing differences of
velocity with differences of time.
Regard the question from the other point of view : assume
that these accelerations are given and we wish to determine the
total change in the velocity in the given period of time. The
total change must be given by the sum of the changes in the small
periods of time; in the first period of -05 sec. the average
acceleration was 26, i, e., the velocity was being increased at the
rate of 26 units per sec. each sec. ; and therefore the change in the
velocity in -05 sec. = 26 x -05 units per sec.
= 1-3 units per sec.
In the successive periods the changes in velocity are -8, 2-9
and 3-1 respectively.
Hence the total change in the velocity over the period -2 sec.
= i-3-f--8+2-9+3'i == 8-1 units per sec., or if the initial velocity
was 28-4, the final velocity was 28-4+8-1 = 36-5. Note that the
acceleration is given by the fraction ^, whilst a small change in
ot
the velocity is of the nature aU, or the total change of velocity
«= sum of all small changes = ^aSt.
We can thus find integrals by working through the processes
of differentiation, but in the reverse order. If a function, expressed
in terms of symbols, has to be integrated, it is an advantage to
transform the rules for differentiation into forms more readily
applicable; the method, however, being entirely algebraic.
If numerical values alone are given, the integration resolves
itself into a determination of an area.
Hence —
Considered from an algebraic standpoint —
Differentiation implies the calculation of rates of change ;
Integration implies the summation of small quantities.
INTEGRATION
117
From the graphic standpoint—
Differentiation is concerned with the measurement of slopes of
curves ;
Integration is concerned with the measurement of areas under
curves.
Just as special symbols are used to denote the processes of
differentiation, so also there are special symbols for expressing the
processes of integrations.
Regarding an integral as an area, it must be of two dimensions,
a length and a breadth; and we have seen in an earlier chapter
(Part I, Chap. VII) that in order to ascertain an area correctly
its base must be divided up into small elements, the smaller the
better, these elements
not necessarily being of
the same length, but all
being small. Thus, to
find the area ABCD-
(Fig. 27) we can suppose
it divided up into small
strips, as EFGH, then
find the area of each of
these and add the results.
The portion EH of the
curve is very nearly
straight, so that EFGH
is a trapezoid, and hence
its area = mean height
X width. Now its mean
6JG
I
F G
FIG. 27.
height FE and GH are practically the same, so that any one of
them can be denoted by y; also the width FG of the strip is a
small element of the base, *'. e., is 8x.
Hence, the area of the strip EFGH — y x 8x, and the total
area between the curve, the bounding ordinates and the axis of x
must equal the sum of all products like y8x, or, as it might be
expressed —
Area = 2yS# (approximately).
However small the width of the strips are made, this sum only
gives the area approximately, but as 8x is diminished the result
approaches the true more and more closely.
Therefore, bearing in mind our previous work on limits, we
can say that the limiting value of 2ySx must give the area exactly.
To this limiting value of the sum different forms of symbols are
n8 MATHEMATICS FOR ENGINEERS
attached, the 2 and S being replaced by the English forms /
and d respectively, so that the area between the curve and the
axis of x = fydx. There is no limit placed to this area in any
horizontal direction, so that the area is not denned by the given
formula.
Hence fydx is spoken of as an indefinite integral.
The x is again the I.V., and the size of the area will depend
on the values given to it. Suppose that when y = AB, x = a,
and when y = CD, x = b ; then the range of x is from a to & if
it is the area ABCD that is considered. Accordingly we can state
that the area ABCD =fydx, the value of this integral being
found between x = a and x = b, or, as it is written for brevity,
fx=b fb
I ydx, or, more shortly still, I ydx, it being clearly understood
J x=a J a
that the limits a and b apply to the I.V., i. e., that quantity
directly associated with the " d."
It is evident that ABCD is a definite area, having one value
fb
only, and thus I ydx is termed a definite integral.
J a
The most convenient method for determining areas (provided
that a planimeter is not handy) is undoubtedly the " sum curve "
method treated in Part I, Chap. VII ; the great virtue of it being
that the growth of the area is seen, and that either any portion
or the whole of the area of the figure can be readily found by
reading a particular ordinate.
In view of the great usefulness of the process of integration
by graphic means, the method is here explained in detail, following
exactly the plan adopted in Part I, Chap. VII.
Graphic Integration is a means of summing an area with the
aid of tee and set square, by a combination of the principles of
the "addition of strips" and "similar figures." An area in
Fig. 28 is bounded by a curve a'b'z', a base line az and two vertical
ordinates aa' and zz'. The base is first divided in such a way
that the widths of the strips are taken to suit the changes of
curvature between a' and z' ', and are therefore not necessarily equal ;
and mid-ordinates (shown dotted) are erected for every division.
Next the tops of the mid-ordinates are projected horizontally on
to a vertical line, as BB'. A pole P is now chosen to the left of that
vertical; its distance from it, called the polar distance p, being a
round number of horizontal units. The pole is next joined to
each of the projections in turn and parallels are drawn across the
INTEGRATION
119
corresponding strips so that a continuous curve results, known as
the Sum Curve. Thus am parallel to PB' is drawn from a across
the first strip; mn parallel to PC' is drawn from m across the
second strip, and so on.
The ordinate to the sum curve through any point in the base
gives the area under the original or primitive curve from a up to
the point considered.
Referring to Fig. 28 —
Area of strip abb' a' = ab x AB
Pole
FIG. 28. — Graphic Integration,
but, by similar figures —
B'a or BA
bin
ab
whence
AB x ab = p X bm
, area of strip , , . , ,
i. e., bm = - or area of strip = px bm
P
i. e., bm measures the area of the first strip to a particular scale,
which depends entirely on the value of p.
, area of second strip
In the same way nm = -
P
120 MATHEMATICS FOR ENGINEERS
and by the construction nm' and bm are added, so that —
area of ist and 2nd strips
en —
P
or — area of ist and 2nd strips — p x en
Thus, summing for the whole area —
Area of aa'z'z = p x zL
Thus the scale of area is the old vertical scale multiplied by the polar
distance ; and accordingly the polar distance should be selected
in terms of a number convenient for multiplication.
E. g., if the original scales are —
i" = 40 units vertically
and i" = 25 units horizontally
and the polar distance is taken as 2", i. e., 50 horizontal units;
then the new vertical scale —
= old vertical scale x polar distance
=« 40x50 = 2000 units per inch.
If the original scales are given and a -particular scale is desired
for the sum curve, then the polar distance must be calculated as
follows —
new vertical scale
Polar distance in horizontal units = , , ,-. — , ,
old vertical scale
E. g., if the primitive curve is a " velocity-time" curve plotted
to the scales, i" — 5 ft. per sec. (vertically) and i" = -i sec. (hori-
zontally), and the scale of the sum curve, which is a " displacement-
time " curve, is required to be i" = 2-5 ft., then —
2*S
Polar distance (in horizontal units) = — = -5
and since i" = -i unit along the horizontal, the polar distance
must be made 5".
Integration is not limited to the determination of areas only;
true, an integral may be regarded as an area, but if the ordinate
does not represent a mere length, but, say, an area of cross section,
the value of the integral will in such cases measure the volume of
the solid.
Our standard form throughout will be for the area of the figure
as plotted on the paper, viz., Jydx, where y is an ordinate and
8x an element of the base, but y and x may represent many
different quantities.
Thus, suppose a curve is plotted to represent the expansion of
INTEGRATION
121
a gas; if, as is usual, pressures are plotted vertically and
volumes horizontally, the ordinate is p and an element of the
base is Sv; hence the area under the curve = \zpdv (if the
J n
initial and final volumes are vt and v2 respectively), and since this
/6s
is of the nature pressure X volume, i. e., 7^3 X (ft)3 or ft. Ibs., the
\7*J
area must represent the work done in the expansion.
To illustrate such a case : —
Work
106560
8 10 la 14- 16 18 SO 2£ 24 26
FIG. 29. — Expansion of Steam.
Example i. — It is required to find the work done in the expansion
of i Ib. of dry saturated steam from pressure 100 Ibs. per sq. in. to
pressure 15 Ibs. per sq. in.
From the steam tables the following corresponding values of p
and v are found : — s
v (cu. ft. per Ib.)
4.44
5-48
7-16
10-50
I3-72
20
26-4
p (Ibs. per sq. in.)
IOO
80
60
40
30
2O
15
By plotting these values, p vertically, the expansion curve is
obtained (Fig. 29) ; this being the primitive curve.
Selecting a polar distance equivalent to 10 horizontal units, we
proceed to construct the sum curve, the last ordinate of which
measures to a certain scale the work done in the expansion. Now
the new vertical scale = old vertical X 10, since the polar distance = 10 ;
and also we must multiply by 144, since the pressures are expressed
122
MATHEMATICS FOR ENGINEERS
in Ibs. per sq. in. and must be converted to Ibs. per sq. ft., so that
the work done may be measured in ft. Ibs.
According to this modified scale the last ordinate is read off as
106560 ; thus the work done = 106560 ft. Ibs.
or, as it would be written in more mathematical language —
T26-4
f
J4
pdv — 106560.
Example 2. — The diameters of a tapering stone column, 20 ft. long,
at 6 equidistant places were measured as 2-52, 2-06, 1-54, i'i5, '80
and -58 ft. respectively.
Find its weight at 140 Ibs. per cu. ft.
29\5
6 8 (O 12 14- IG IQ 20
FIG. 30. — Problem on Stone Column.
The volume will be obtained by plotting the areas against the
length and summing. Now the area of any section = -dz, and the
total volume will be the sum of the volumes of the small elements
into which the solid may be supposed to be divided.
/"20 pO^
Thus the volume = J 0 Adi = J Q ~d*.dl.
and the weight = 140 I -d2.dl.
J o 4
Since - is a constant multiplier, it can be omitted until the end,
for its effect is simply to alter the final scale ; hence a constant factor
before integration remains so after.
Hence the weight-
= 1097 j
d*.dl.
INTEGRATION
123
The integral will be of the standard form if for dz we write y and
if for / we write x, so that we see that ordinates must represent d2 and
abscissae lengths, and hence the table for plotting reads : —
I
o
4
8
12
16
20
y or d2
6-34
4-24
2-36
1-32
•64
•336
Plotting these values and thence constructing the sum curve (see
Fig. 30), we find the last ordinate to be 47-15, and this is the value
f20
of J Q d*.dl.
[20
Weight = 109-7 1 dz.dl = 109-7x47-15 = 5180 Ibs.
Application of Integration to " Beam " Problems. — At an
earlier stage (see p. 38) it has been demonstrated that the shear
at any point in the length of a beam loaded in any way whatever
is given by the rate of change of the bending moment in the
neighbourhood considered, this being the space rate of change.
Conversely, then, the bending moment must be found by summating
the shearing force ; and hence, if the shear curve is given, its sum
curve is the curve of bending moment.
In the majority of problems the system of loading is given,
from which the curve of loads can be drawn. Then, since the
shear at any section is the sum of all the forces to the right or
left of that section, the sum curve of the load curve must be the
shear curve; continuing the process, the sum curve of the shear
curve, i. e., the second sum curve from the load curve as primitive,
is the curve of bending moment and the fourth sum curve is the
deflected form.
Expressing these results or statements in the notation of the
calculus ; L, S and M being the respective abbreviations for loading,
shear and bending moment —
S =fLdx
M =f$dx =f(fLdx)dx = ffL(dx)z
[ff~L(dx)z being termed a double integral]
and the deflection y = ff M.(dx)2 or ffff L(dx)*.
If the loading is not uniform, but continuous, the summation
must be performed graphically. [The link polygon method largely
used obviates half these curves, e. g., the link polygon for the loads
gives at once the curve of bending moment.]
124
MATHEMATICS FOR ENGINEERS
Example 3. — The loading on a beam, 24 ft. long, simply supported
at its ends varies continuously, as shown in the table. Draw diagrams
of shearing force and bending moment, stating clearly the maximum
values of the shearing force and the bending moment.
Distance from one end (ft.)
o
4
7
10
12
J4
17
20
24
Load in tons per ft.
•44
•58
•86
1-06
I-I
i -06
•86
•58
'44
The curve of loads is first plotted, as in Fig. 31.
By sum-curving this curve, we obtain the curve of shearing force,
although no measurements can be made to it until account has been
taken of the support reactions.
4 68 1C 12' ' 14 >6 <8
FIG. 31. — Problem on Loaded Beam.
To find the reactions at the ends : We know that these must be
equal, since the loading is symmetrical, each reaction being one-half
of the total load. Now the last ordinate AB of the sum curve of the
load curve is 19; thus the reactions are each 9-5. Bisecting AB, or,
in other words, marking off a length AC to represent the reaction
at A, we draw a horizontal, and this is the true base line for the curve
of shear; any ordinate to the curve of shear from this base giving
the shear at the point in the length of the beam through which the
ordinate is drawn.
We observe that the shear changes sign and is zero at the centre
of the beam; we can conclude from this that the bending moment
must have its maximum value at the centre, since shear — rate of
change of bending moment, and if the shear is zero, the bending
moment must have a turning value.
By sum-curving the shear curve from CD as base, the resulting
curve is that of bending moment.
INTEGRATION
125
It is well carefully to consider the scales, for it is with these that
difficulties often arise.
The scales given here apply to the original drawing, of which
Fig. 31 is a reproduction somewhat under half full size.
For the length i in. = 2-5 ft.
For loads i in. = -4 ton per ft.
Polar distance for the first sum curve, i. e., the curve of shear —
= 4 ins. = 4X2-5, or 10 horizontal units.
Hence the scale of shear = -4 x 10, or 4 tons to i in.
B
FIG. 32. — Shearing Force and Bending Moment on Ship's Hull.
Polar distance for the second sum curve = 4 ins. = 10 horizontal
units.
Hence the scale of bending moment = 4 X 10 = 40
or i in. (vertically) = 40 tons. ft.
Reading according to these scales —
The maximum shear = 9-5 tons ^
and the maximum bending moment = 68 tons, ft. J
Example 4. — In Fig. 32 AAA is the curve of weights or load
distribution, and BBB the curve of buoyancy or upward water thrust
126 MATHEMATICS FOR ENGINEERS
for a ship whose length is 350 ft., the scale of loads being indicated
on the diagram.
Draw diagrams of shearing force and bending moment on the hull
of the vessel and measure the maximum values of these quantities.
It is first necessary to construct the curve of loads to a straight
line base, and to do this the differences between the curves AAA and
BBB are set off from a horizontal, taken in our case below the original
base line.
In this way the curve of loads LLL is obtained, the scale being
shown to the left of the diagram.
By sum-curving this curve, the curve of shear SSS is obtained ;
the polar distance (not shown on the diagram) being taken as 50 horizontal
units, so that the scale for the shear is 50 times the scale for the loads.
Sum-curving the curve SSS, the curve MMM, that of bending
moment, is obtained (again the polar distance is 50 horizontal units).
Sectio'ns such as K, where the upward thrust of the water balances
the downward force due to the weights, are spoken of as water-borne.
Reading our maximum values according to the proper scales, we
find them to be —
Maximum shear = 246 tons ~\
Maximum bending moment = 14,300 tons ft. /
It should be noted that the last ordinate of both the shear curve
and the curve of bending moment is zero ; these results we should
expect since the areas under the curves AAA and BBB must be
equal, so that the shear at the end must be zero, and also the moments
of these areas must be alike.
[In practice the maximum bending moment is found by such a
formula as—-
Weight X length
Maximum bending moment = 7? —
Constant
the constant for small boats being between 30 and 40, and for larger
between 25 and 30.]
The Coradi Integraph. — A brief description of the Integraph,
an instrument devised to draw mechanically the sum curve, can
usefully be inserted at this stage.
It consists essentially of a carriage running on four milled
wheels A (Fig. 33), a slotted arm C carrying the tracer B which
is moved along the primitive curve, and the arm D which carries
the pencil E which draws the sum curve.
As B is moved along the primitive curve, the slotted arm C
slides about the pins G and P, thus altering its inclination to the
horizontal. A parallel link motion ensures the movement of E
parallel to the instantaneous position of C, the sharp-edged wheel F
assisting in guiding the tracer bracket.
INTEGRATION
127
The principle of the instrument is not difficult to understand,
and can be explained in a very few words.
The pole is at P, and the tops of the mid-ordinates are pro-
jected to the vertical through G by the horizontals like BG;
parallels are then drawn to PG by the pencil E, the motion being
continuous.
The polar distance can be varied as desired, by altering the
position of the bracket carrying the pin P along the horizontal
arm; and if an extremely small polar distance is found to be
advisable, the pin H may be utilised instead of G.
Rules for Integration of Simpler Functions. — Since
integration is the reverse of differentiation, many functions can be
integrated by reversing the order of the steps in differentiation.
Integration of powers of x.— The first rule given in the
work on differentiation of functions was —
d
— xn = nxn~v.
dx
128 MATHEMATICS FOR ENGINEERS
To change into the integration form, we transpose -5- : the " d"
on the one side becomes f on the other side, to indicate the change
differencing to summing, and the " dx " occurs on the top line of the
other side of the equation.
Thus — xn = fnxn~1dx
or fxn~1 dx = - xn-\-C
%
the reason for the presence of the constant term C being explained
later.
It is a trifle simpler to write n in place of n—i, and therefore
«+i in place of n, so that —
Whereas, when differentiating a power of the I.V., the power
was reduced by i in the process of differentiation ; when integrating,
the power is increased by i.
fa — <y-5 — tiv^
-g" dxx
whereas fx5dx = ^ x6-\-C.
A special case occurs for which the above rule does not apply :
for, let n = — i, then fx~ldx should, according to the rule just
given, be - x°, but to this fraction no definite meaning can be
assigned.
We know that -3- log x = - = x~*
dx x
fx-1 dx = log x+C
or, as it is sometimes written —
dx ,
A constant multiplier before integration remains as such after;
thus faxndx = — T— %W+1+C.
Also an expression composed of terms can be integrated term
by term, and the results added.
INTEGRATION 129
Thus, f (axn-\-b) dx can be written —
faxndx -\-fbdx
i. e., fax*dx-\-fbx?dx f or x° = i
its value being -—xn+l -\-bx+C.
Note. — Differentiation of a constant term gives zero, but the
integration gives that constant multiplied by the I.V.
The reason for this will be apparent if we consider the state-
ments from the graphical standpoint. The curve representing the
equation y = b is a horizontal straight line, and therefore the
slope is zero (i.e., -v- = o) ; but the area under the curve = the
area of a rectangle = base x height = xxb (i. e.,fbdx = bx).
Exponential Functions. — We have already proved that
dx
dex
—- — e* (See p. 47) ; then by transposition of d and dx to the
other side of the equation we obtain the statement ex = j
and since -, corresponds to f we may write this as fexdx = ex-\-C.
Thus if we either differentiate ex or integrate it we arrive at the
same result ; and ex is the only function for which the differential
coefficient and also the integral are the same as the function itself.
Carrying this work a step further, let us consider the integration
of ebx, and hence ax : —
Now f-aebx = abebx .*. faebxdx = ?J*+C.
dx b
To avoid confusion as to the placing of a and b we must reason
in the following manner : The a is a constant multiplier of the
whole function, and therefore remains so after integration; the b
multiplies the I.V. only; and thus differentiation would cause it
to multiply the result, whereas after integration it becomes a
divisor. Great attention should be paid to the application of this
rule, for unless care is exercised mistakes are very apt to creep in.
Example 5. — Find the value of -^(15**— 7*'9+83) and also of
130 MATHEMATICS FOR ENGINEERS
Differentiating the first expression —
d.
Integrating the second given expression —
4 --i + i
= i5t*-7f»+C.
Notice that although a function has been differentiated and the
derivative integrated, the final expression is not exactly the same as
the original, the constant term being represented only by C, where C
may have any value. Further reference will be made to this point
on p. 137.
Example 6. — If pv1'32 = C, find the value oi/pdv.
To express p in terms of v —
fpdv = fCir-1-** . dv = Cfv~l-3Zdv
(K being any constant)
= CX- — u-l-38+1
-1-32+1
= — —
•32
This result can be written in a slightly different form, if for C we
write its value pv1'32; then —
F., pvl'**xir-** . v
ipdv = — — — \-K
•32
pv
— — = — 3*
•32 — a
Exa'.-nple 7.— Find fpdv when pv = C.
In this case — p = Cv~*
fpdv = fav - C/^ = C log v + K
= pv log v
Example 8. — Find the value of
Note that 17 is a constant multiplier throughout; 2 multiplies the
I.V. and therefore appears as a divisor after integration ; also the
power of e remains exactly the same.
INTEGRATION
Example 9. — Find the value of f(^oe&v+v5-*)dv.
f(4oe'*>+vs-*)dv = f^oe&vdv+fu5-*dv (separating the terms)
-e
5
131
Example 10. — Find an expression for faxdx, and apply the result
to determine the value of /i2 x
From our previous work we know that —
d
dx
ax = ax . log a.
faxdx = .— -.*
* log a
Hence fiz x 5txdx = (i2X- X .— ^— X 54*)+C
loge 5 = 1-609
= 1-864 X54g+C.
Afofe. — It would be quite incorrect to multiply 1-864 by 5 and
express the result as g-^2*x.
Alternatively, the result might have been arrived at in the following
manner —
= 12 x, H— X(54)*+C
log 625 VJ '
= g; — X54ar+C (log 625 = 6-44)
= 1-864 X54*+C.
Exercises 12. — On Graphic Integration.
1. The acceleration of a slider at various times is given in the
table. By graphic integration obtain the velocity and displacement
curves to a time base, indicating clearly your scales.
j Time .
o
•008
•016
•02
•028
•036
•044
•048
•06
! Acceleration
o
75
87-5
87-5
87-5
87-5
87-5
83
O
•068
•072
•084
•10
•108
•12
78
85
87-5
87-5
83
O
132
MATHEMATICS FOR ENGINEERS
2. An acceleration diagram on a time base has an area of 4-7
sq. ins. The base of the diagram is 2-5 ins. and represents 25 sees.
The acceleration scale is i in. =3 ft. per sec.2. If the velocity at the
beginning is n ft. per sec., find the velocity in ft. per sec. at the end
of the 25 sees.
3. A rectangular barge is loaded symmetrically in still water.
The curve of loading is a triangle with apex at the centre, and the
curve of buoyancy is a rectangle. Draw diagrams of shearing force
and bending moment on the barge.
4. The curves of loads for a ship 350 ft. long is as given in the
table. Plot this, and by graphic integration obtain the curves of
shearing force and bending moment.
Distance from one end (ft.)
0
7
10
35
56
84
102
load (tons per foot) .
0
•3
o
-2-6
-3-i
-2-3
O
112
133
161
196
210
237
260
280
315
330
350
1-6
2-3
3-15
5'2
57
0
-4-5
-4-95
o
•9
0
5. The table gives the values of the pressure and volume for the
complete theoretical diagram for a triple expansion engine.
V
o
I
2
4
6
8
10
12
p
240
240
I2O
60
40
30
24
20
Find the initial pressure in each cylinder in order that the work
done per cycle may be the same for each.
(Hint. — Divide the last ordinate of the sum curve into three equal
parts, draw horizontals through these points of section to meet the
sum curve, and from these points of contact erect perpendiculars to
cut the expansion line.)
6. A body weighing 3000 Ibs. was lifted vertically by a rope, there
being a damped spring balance to indicate the pulling force F Ibs. of
the rope. When the body had been lifted x ft. from its position of
rest, the pulling force was automatically registered as follows : —
X
0
20
40
65
75
95
no
140
F
8000
7950
7850
7500
7400
6800
6400
4000
Find the work done on the body when it has risen 80 ft. How
much of this is potential energy and how much is kinetic energy ?
Find, also the work done when it has risen 140 ft.
INTEGRATION
133
7. The current from a battery was measured at various times,
with the following results : —
Time (hours) .
0 I
3
6
9
10
12
14
15
Current (amperes)
25 28
37
39'5
32
29
24
25-3
2?
If its capacity is measured by fCdt, find the capacity in ampere
hours.
8. The following are the approximate speeds of a locomotive on a
run over a not very level road. Draw a curve showing the distance
run up to any time.
Time (mins. and sees.)
0
I
2-15
6.15
9.22
11-45 14.26
16.35
20.52
Speed (miles per hr.)
0
6
10
18-2
22-8
25-5 28
29-2
28-6
9. The load curve at a large central station can be constructed
from the following data : —
Time (hours) . . ,
0
I
2
3
4
5
6
14
7
7-5 8
9
10
ii
Load (1000 amperes)
3'5
i
I
2
I
6-4
i7
17-8 16-4
ii'3
8-7
8-2
12
I
2
3J4
5 i 5'5
6
7
8
9
10
II j 12
|7-8
8
7-6
8-7 12-5
19 23-3
21
12-4
ii
10-5
9-6
9 ! 6
Find the total number of ampere hours supplied in the 24 hours.
10. The velocity of a three-phase electric train, with rheostatic
control, at various times, was found as in the table : —
Time (sees.) . . . | o
26-6
66-6
80- 1
99
Velocity (ft. per sec.) . o
40
40
37-3
0
Draw the space-time curve and find the total distance covered in
the 99 seconds.
On the Integration of the Powers of x and of the Exponential Functions.
11. What is the significance of the symbols f and dx in the
expression fxzdx ?
Integrate, with respect to x, the functions in Examples 12 to 27
12. 4*1-". 13. 70-15.
14.
15.
16. e'™.
17. *»-— +14.
x
18.
19.
134 MATHEMATICS FOR ENGINEERS
20. i2*~. 21. -g^. 22. -i7e". 23.
24. 2.54*--i«-8.2*-i++i-i3. 25.
26. ^.«*_*--«3+ 27. -94*'18 cos «-
Find the values of the following —
2B.fv5dv. 29. /"-". 30. /35^. 31. /g"-«<fc
32. //xfo when ^y1'17 = C. 33. /i4 X 2*^5.
-8)d*. 35. /3- 1'««. 36. 17 /^g.
J 0 r
38. $-z-2(dt)z. 39. 2-ix~5dx
40. Solve the equation -/- = — w -.
rfy y
41. In connection with the flow of air through a nozzle, if x is the
distance outwards from the nozzle and v is the velocity there, v oc -1 .
Also SA (an element of area of flow) = ~K8x Vx. The added momentum
for the small element considered = SM = v8A. Show that M the
D
total increment to the momentum, can be written C— —7= where C
vx
and D are constants.
Trigonometric Functions. — We have previously seen that
the derived curve of either the sine curve or the cosine curve is the
primitive curve itself transferred back a horizontal distance of
one-quarter of the period. Conversely, then, we may state that
the sum curve of either the sine or the cosine curve is the curve
itself moved forward for a distance corresponding to one-quarter
of the period. In other words, integration does not alter the form
of the curve. Taking the case of the sine curve as the primitive,
we see, on reference to Fig. 34, that if this curve is shifted forward
for one-quarter period the resulting curve is the cosine curve
inverted ; or expressing in algebraic language, whilst the equation
of the primitive curve is y = sin x, that of the sum or integral
curve is y = —cos x. Thus, fsmxdx = — cos x. In like manner
it could be shown that feosxdx = sin*. For emphasis, the dif-
ferentiation and the integration of sine x and cosine x are repeated
here —
-T- sin x = cos x /sin* dx = — cos #4-C
dx
-T-COS x — —sin x /cos* dx = sin x-\- C.
dx J
INTEGRATION
135
Note. — When differentiating the cosine the minus sign appears in
the result ; when integrating the sine the minus sign appears ; it is
important to get a good grip of these statements, and the con-
sideration of them from the graphic aspect is a great help in this
respect.
To extend the foregoing rules —
fcosxdx = sin x-\-C
/cos (ax+b)dx = sin (ax+b)+C
fsinxdx = —cos x-\-C
/sin (ax+b)dx = — cos (ax+b)+C.
1
•75
•5
•25
O
•25
•5
•75
1
^
\/
J?
N
\
/
s~
A
\
. //
-si
n..
rr
/
/
\
Y
/
I
\
\
/
JG
%
1
\
/
*
24\
-y
-a
/
is.<
r\
\
/
/
\
y
/
\
V
s
A
V
s
/
FIG. 34.
Thus the angle remains the same after integration just as it
would after differentiation, but the constant multiplier a of the
I.V. becomes a divisor.
To integrate sec2 -AC with regard to x we call to mind the
differentiation of tan x, viz., , tan x = sec2 x. Accordingly
/sec2 xdx = tan x+C.
Extending this to apply to the more general case —
/sec2 (ax+b)dx - -1 tan (ax+6)+C
In like manner —
/cosec2 (ax+b)dx = — - cot (ax+6)-{-C.
I36 MATHEMATICS FOR ENGINEERS
Other two standard integrals are added here, the derivation of
which will be considered in the next chapter.
/tan* dx — — log cos x+C
fcotx dx = log sin x+C.
To verify these we might work from the right-hand side and
differentiate. Dealing with the former —
d , , . _ _ d , if u = cos %
dx^ ° dx ° whence
d log u du du
— _ _ v — — Qin v
- 1 S\ 7 7 ' Oil! Jv
du dx dx
*
i
= — x— sin x
u
sin x
= - = tan x
cos*
/tan* dx = —log cos x+C.
Example n. — Find the value of f(5~ sin^t)dt.
— f$dt— fsin^tdt
= ^— ~X —cos
V4
= 5'+- cos 4*+C.
4
Example 12. — Evaluate /sin (5— ^t)dt.
/sin (5—4t)dt = --x -cos (5— 4/)+C = ] cos (5— 40 + C.
4 4
Example 13. — If a force P is given by P = 36-4 sin (1005— -62),
find the value oifPds.
= —-364 cos (1005 — 62) + C.
Example 14. — Find the value of
12 cos (4-
The expression E = 7-25*— 3s~1+i5s-8+ 12 cos (4 — 35)
= (7-2X^)-3logs+i;|si-8+(i2X-^sin (4-35))+C
= I-44S5— 3 Jog 5 + 8-33S1-8— 4 sin (4— 3
fPds =/36'4 sin (loos— -62)^5 = — ^X —cos (1005— -62) + C
INTEGRATION 137
Example 15. — If R = n sec2 (3—4-71;), find /Rdv.
fRdv =/n sec2 (3 — 4-?v)dv = - — tan (3— 4-7t>)+C
= —2-343 tan (3—4-
Exercises 13. — On Integration of Trigonometric Functions.
Integrate, with respect to x, the functions in Nos. i to 10.
1. 3 sin 4*. 2. —5-18 cos (3 — 3^). 3. 7 sec2 (^3 — x).
4. x— °12— -14 cos (-05— -117*). 5. 05-4*4-5 sjn ^b+ax).
6. 9-45 sin 8^. 7. —3-08 sin 2(2-16*— 4-5).
8. 9^^+^-1-83 tan x.
9. 4-27 sin (--- — J + -2 cos gx— 4#1-74+32*+5.
10. 2 sin2 #—2-91 sin ( — 3'7#)+2 cos2 #—14-2 cosec2 ----- .
11. The acceleration of a moving body is given by the equation —
a = —49 sin (jt— -26).
Find expressions for the velocity and the space, the latter being
in terms of the acceleration.
.«« Tf dzx o o / , cos 20\ _ , ,, , dx
12. If T-4 = 47c2wV^cos 0-i J, find the values of , and x.
(x is a displacement of the piston in a steam-engine mechanism.)
13. Find the value of /b^+cos (y7—7"2p)}dp.
14. If v — 117 sin 6^—29-4 cos 6t, find the value offvdt.
-«
Indefinite and Definite Integrals. — The integrals already
given, although correct, are not complete. If "an integral is to
denote an area some boundaries must be known ; and nothing was
said about the limits to be ascribed to x (or s, as the case might
be) in the foregoing, so that we were in reality dealing with
indefinite areas or integrals. To indicate that a portion of the
area may be dispensed with in certain cases (when the boundaries
are stated) a constant C is introduced on the R.H.S. of the equation,
x*
i. e.,fx3dx would be written +C.
As soon as the integral, and therefore the area, is made definite
it will be observed that C vanishes.
138 MATHEMATICS FOR ENGINEERS
If fx3dx is to equal -x*+C, -, (-#4+Cj should equal x3 ; and
this is the case for —
d
-=-
dx
^A 4 c\ ^
dx\4 J dx £
(C being independent of x).
Cf. Example 5, p. 129 ; the constant in that case being 83.
It is therefore advisable to add the constant in all examples
on integration; in many practical examples the determination of
the value of the constant is an important feature, and therefore
its omission would invalidate the results obtained.
In the list of a few of the simpler standard integrals collected
together here for purposes of reference and by way of revision
the constant is denoted by C.
f(axn+b)dx a
faxndx
Jbdx
[dx
J x
[ dx
J ax+b
= bx+C
= log x+C
= ^ log («*+&) +C
faebxdx
- ab^+c
fe*dx
= e*+C
faFdx
= rog*x«*+c
/sin (ax-{-b)dx
- cos (ax+b)-\-C
f sin xdx
= —cos x-\-C
/cos (ax-\-b}dx
f cos xdx
/sec2 (ax-\-b}dx
f sec2 xdx
/cosec2 (ax-\-b}dx
= - sin (ax-}-b)-{-C
&
sin x-\-C
- tan (a^
= tan
-- cot (ax+b)+C
d
INTEGRATION 139
/ cosec2 xdx = —cot x-\-C
/tan (ax+b)dx = — log cos (ax-}-b)-\-C
8
/ tan xdx = —log (cos #)+C
/cot (ax+b)dx = log sin (ax+b)-{-C
f cot xdx = log (sin A;) +C.
Method of Determining the Values of Definite Integrals.
— We may regard the area of a closed figure as the difference
between two areas, viz., all the area to the left, say, of one boundary,
minus all the area to the left of the other parallel boundary.
Hence to find the value of a definite integral, the value of the
integral must be found when the I.V. has its higher limiting value,
and from this must be subtracted its value when the lower limiting
value is substituted for the I.V.
fx2dx = -x3-\-C is an indefinite integral, but if to C we give
a definite value, it becomes definite and unique.
/•4
Thus I x2dx is a definite integral, because the limits to be
J 2
applied to x are indicated.
To evaluate it —
We know that fxzdx = -x3-\-C.
The value of this integral when x = 4 is — -f-C
o
and the value of this integral when x = 2 is — |-C
the constant being the same in the two cases.
The difference = -^+C - -+C = ^
V 3 / \3 / 3
meaning that if the curve y = xz were plotted, and the area between
the curve, the x axis and the ordinates through x = 2 and x = 4
found, its value would be i8| sq. units.
It will be noticed that C vanishes, and hence when dealing
with definite integrals it is usual to omit it altogether.
) MATHEMATICS FOR ENGINEERS
/"4 /#3\4
For brevity, I x2dx is written I — ) , which on expansion
.'2 \3/2 r
reads —
__ e 56
3 3/'* '3'
Example 16. — Find the value of the definite integral I
* -i
f V**d* = f^ g3*Y4 = 4
J .,' \3 /-i 3
3 v
= | (3-3201-1-3499)
= 2-62 <).
Example 17. — Evaluate the definite integral
V
f2
/ (50034^+7) rf^
.'
5
I (50054^+7) dx = (- si
•> o H
sn
7?r
= '— or ii.
2
Example 18. — Find the value of
/
J o
The expression
= n-2
4
INTEGRATION
141
Notice that no cancelling takes place, beyond that concerning the
constant multiplier 5, until the values (4 and 2) have been substituted
in place of x. In other words, it would be quite wrong to say —
c
4/i
Example ig. — The total range of an aeroplane in miles can be
C Wt
obtained from the expression — / dq where m = pound-miles
per Ib. of petrol, and q =
loading at any time
initial loading
Taking q = -6 and m = 4000, find the total range.
= — m (log q— log i)
= — m log q.
Now if q = -6, log q = 1-4892 = —-5108.
Hence the range = 4000 x -5108 = 2043 miles.
— c-
A F B O L
FIG. 35. — Proof of Simpson's Rule.
Proof of Simpson's Rule for the Determination of Areas
of Irregular Curved Figures. — This rule, given on p. 310 of
Part I, states that—
length of one division of the base f first + last „ ordinate +
= — — —\42 even ordmates -f-
[22 odd ordinates.
142 MATHEMATICS FOR ENGINEERS
It is now possible to give the proof of this rule.
Let us deal with a portion of the full area to be measured,
such as ABCD in Fig. 35. Let the base AB = 2c.
Let the equation of the curve DEC be y = A4-B#+C#2, so
that DEC is a portion of some parabola.
We can assume that the origin is at F, and therefore the abscissae
of D, E and C are — c, o and +c respectively.
Hence AD = yx = A+B(-c)+C(-c)2 = A-Bc+Cc2
FE = y2 = A+B(o)+C(o)2 = A
BC = y3 = A+B(c)+C(c)2 = A+Bc+Cc2.
f+e
Now the area ABCD = I ydx
J -c
= i+
/ -
. . Be2 , Cc3 , . Be2 , Cc3
—. A /»_!_ _L A r _
"^ 3~+ "2 3"
A ,
- 2AC+
= -{6A+2Cc2}
= -{A-Bc+Cc2+4A+A+Bc+Cc2}
Imagine now another strip of total width 2c added to the right
of BC; the double width being chosen, since there must be an
even number of divisions of the base.
Then if GH = y4 and LK = y6
Area of BLKC = -
O
or area of ALKD = -{
O
= ~{
If a strip of width = 2c is added to the right
Area = Z
INTEGRATION 143
Or, in general —
Area = -{first+last+42 even+22 odd}.
o
Exercises 14. — On the Evaluation of Definite Integrals.
Find the values of the definite integrals in Nos. i to 7.
i. rw 2. /•"* 3.
./ 1-02 J 1-7 W
«•
4. P5-i sin -26d». 5.
p.7
6. I s-2*'1^. 7.
-- .
I x-"dx
8. The change in entropy of a gas as the absolute temperature
f775 dr
changes from 643 to 775 is given by I -85—. Find this change.
* 643 T
IT
9. If H = ^ 1 2 sin 6d6, find the value of H.
pJ o
10. The average useful flux density (for a 3-phase motor)
i fis*'
= B = - Bmar sin 6d6. Find B in terms of ~Bmax-
itJ _
12
11. Express sin at cos bt as the sum of two terms and integrate
with regard to /. If a is -=? and b is 30, what is the value of the
integral between the limits o and T ?
12. If h = • find h.
g J RI r3
13. Given that EI^ = ^-^-P*. Also that ^ = o when
a^r2 22 d#
x = I, and y = o when .# = o and also when .*• = /; find the value
of P and an expression for y.
wx2 M d'ty dy
14. If M = - , T = ET~-,, ~-= o and also y — o when x = I,
2 I dx2 dx
find an expression for y.
15. Given that M - -(--*2)-K, ^ = f? Also ^ = o when
2 \4 / IE dx2 dx
I I
x = -, and y = o when x = ±- ; find an expression for y. (The case
of a fixed beam uniformly loaded.)
16. Find the value of J 0(/# -**)***.
144 MATHEMATICS FOR ENGINEERS
17. Evaluate 3 I jX(lz— zlx-\-xz)dx, an integral occurring in a beam
J "2* »
problem.
18. If Q = / qdx and q = — ; — ; — -wx, find Q, the total horizontal
J «^ i+sin <t>
thrust on a retaining wall of height h, w being the weight of i cu. ft.
of earth, and <J> the angle of repose of the earth.
19. Find the area between the positive portion of the curve
y = 3#— 4#2+n and the axis of x, and compare with the area of the
surrounding rectangle.
/15-8
pdv when pv1-37 = 594.
4*6
d i/ ^ d'V
21. If ~z = 6*1'4 — \ : -,- — 10-5 when x = i, and y = 14 when
Q/X X dsG
x = 2, find an expression for y in terms of x.
22. Evaluate / • — - — fiXf
J 1-47 13 — $X
23. Find the value of n, given by the relation n = I - — -. — .
J n t
r ,'r
24. The total centrifugal force on a ring = / 1 - '- — ^ — ; find
J E2 -t^i
an expression for the force.
25. The area of a bending moment diagram in a certain case was
f'/i «3\
given by — J [-at -- -,)da', find the value of this area.
26. H, the horizontal thrust on a parabolic arch,
i
Find an expression for H.
27. The work done by an engine working on the Rankine cycle
with steam kept saturated = | l— dr.
j\T
Find the work done if the temperature limits are 620° F. and
800° F. (both absolute), and L = 1437—71-.
28. Evaluate
T2.4
J [>-5*— sin (25—
29. Find the value of n, the frequency of transverse vibrations of
a beam simply supported at its ends and uniformly loaded with w
tons per foot run, when the equation of the deflected form is —
y =
and
Ci ,
ydx.
INTEGRATION 145
30. From Dieterici's experiments we have the following relations —
If s — specific volume of liquid ammonia
and c = specific heat of liquid ammonia
then for temperatures above 32° F. —
c — i-n8 + -ooii56(£— 32)
and s = / cdt.
J o
Find s when * = 45° F.
31. If p = ^-j\s2-xz}dx, Q being the leakage of fluid past a
WJ»J o
well-fitting plug, find its value.
32. The total ampere conductors per pole due to the three windings
/- / A • J
in a railway motor — CV2 / — Ai sin -rdx.
2 Jo* I
Evaluate this integral.
33. For a viscous fluid flowing through a narrow cylindrical tube
of radius v, the quantity Q is given by the formula —
O — prc/fo2
1* 2/*
where /* is the coefficient of viscosity.
Find the value of Q.
CHAPTER VI
FURTHER METHODS OF INTEGRATION
BY the use of the rules enumerated in the previous chapter it
is possible to perform any integration by a graphic method and
the integration of the simpler functions by algebraic processes.
Whilst the graphic integration is of universal application, it at
times involves much preliminary arithmetical work, which it is
tedious to perform, so that it is very frequently the better plan to
resort to a somewhat more difficult, though shorter, algebraic
method. For the more complex functions, then, a choice has to
be made between the two methods of attack ; the fact being borne
in mind that only in cases where definite integrals are concerned
does the graphic method of integration compare favourably with
the algebraic.
It is therefore advisable to introduce new processes and artifices
to be employed for the algebraic integration of difficult functions;
and whilst it is not absolutely essential that all these forms should
be remembered, it is well that the various types should be
considered, so that they may be recognised when they occur.
It is impossible to deal here with every kind of integral likely
to be encountered ; all that can be done is to develop the standard
forms which cover a wide range, and to leave them to suggest
forms for particular cases.
Integration by the Aid of Partial Fractions. — Many com-
plex fractions can be split up into simpler or partial fractions, to
which the simple rules of integration may be applied. Thus if we
Q/£ OQ
are asked to integrate, with respect to x, the fraction — 2_ ~ ,
we soon discover that we are unable to perform this operation
with only the knowledge of integration acquired from the previous
chapter.
If, however, we break the fraction up, in the manner explained
in Part I, Chap. XII, we find that the integration resolves itself
into that of two simple fractions.
146
FURTHER METHODS OF INTEGRATION 147
Thus— ,8*~3? o = - — h-4- (see Part I, p. 453).
2— x— 4 2x— 7
Hence- f**^?** = f-2- d*+ l
J 2x2— 15^+28 J x— 4 J
2#— 7
= 2 log (*— 4) +4 log (2*— 7) + log C
= log (*-4)2+log (2*-7)2+log C
= log {C(*-4)2(2*-7)2}.
[Note that log C may be written to represent the constant in
place of C alone ; and it can then be combined with the other logs.]
/dx
~i 2
xz—a2
i A B
*2_a*- (*_a) +
Equating numerators —
Let ^ = a, then i = A(2a)+o
and A = — .
20
Let x = —a, then I = o + B( — 2a)
and B = -2J«
i =JL(_J_. _i_l
x*—az 2a\x—a x-\-a'
_, /" dx i f f dx f dx 1
Hence — -= „ = — i / —
J x2—az 2.a\-' x—a J x+a)
= ^{log (^— «)— log (#+a)+log C}
= ^;log
This is a standard form.
A rather more general result may be deduced from it.
Example 2. — To find J ,
Let (x+a) = X
- f -—T= -. «" / g
A Ilcll"" r / . \ A v a ^— !•*-.-» •• A , i** ~r~ fc* i j
rf^r
Explanation.
x+a = X
(X+6)
C(^+a— &) and thus for dx we may
write rfX.
148 MATHEMATICS FOR ENGINEERS
Integration by the Resolution of a Product into a
Sum. — A product cannot be integrated directly; but when the
functions are trigonometric the product can be broken up into a
sum or difference and the terms of this integrated.
Before proceeding with the work of this paragraph the reader
would do well to study again pp. 273 to 286, Part I.
Example 3. — Find the value of /4 sin 5^.3 cos2ldt.
4 sin 5^.3 cos 2t = 12 sin 5^ cos 2t
= 6 X 2 sin 5* cos 2t
= 6{sin 7/+sin 3*} (cf. p. 286, Part I).
Hence —
/4 sin 5^.3 cos 2tdt = 6[f sin jtdt+fsin $tdi\.
= 6J — cos jt — cos 3/+C J
= 6C — cos 7^—2 cos
Example 4. — Find /sin2 xdx.
cos 2.x = i — 2 sin2 x, so that —
sin2 x = i -cos 2* (cf> p 2g
= -5* — -25 sin2*+'5C.
Example 5. — Find f ta,n2xdx.
We know that sec2 x = i + tan2 x.
.'. fta,n2xdx =f(secz x—i) dx —fseczxdx—fi dx.
= tan x—x-\-C.
Integration by Substitution. — At times a substitution aids
the integration, but the cases in which this happens can only be
distinguished after one has become perfectly familiar with the
different types.
y- is a type to which this method applies.
In this fraction it will be observed that the numerator is exactly
the differential of the denominator. Hence if u be written for the
FURTHER METHODS OF INTEGRATION
149
denominator, the numerator may be replaced by du, so that the
integral reduces to the simple form I — , i. e., log w+log C.
J u
For if
or
Hence
u = axz+bx-\-c
du
-j- = 2ax-\-b
dx
du = (2ax-\-b)dx .
((^x+b^x = fdu
J axz-}-bx-\-c J u
= log w+k>g C
= log Cu
= log C(ax*+bx+c).
In many cases integration may be effected
by substitution of trigonometric for the algebraic
functions ; and Examples 6 to 10 illustrate this
method of procedure.
Example 6. — To findyVa2—
-X
FIG. 36.
Let
then
x — a sinw, as illustrated by Fig. 36
a2— x2 = a2 — a2 sin2u = a2(i — sin2w) = a2 cos2u
and Va2— x* = a costt, as will be seen from the figure.
A , dx d(a sinw)
Also -,- — -*—s - = a cosu
du du
i. e., dx = a cosw . du.
J'-\/a2—x2 dx — fa cosu . a COSM du
= a2f cos2 u du
a?
= —f(i + cos2ii)du, since cos 2A = 2 cos2 A— i
= — ( u-\ — sin 2M + CJ.
Although this result is not expressed in terms of x, it is left in
form convenient for many purposes.
To express the result in terms of x —
x x
sin u = -, so that u = sin-1 -
a a
and also
cos u
/a2 _ x%
= \f - = — .
» n&
150 MATHEMATICS FOR ENGINEERS
Hence - sin aw — sinw cosu = - x - Va2— x2
•z a a
az-x*dx = —X sin-1 - +(TX^2 Va2-*2 +K
V 2 •/ \ 2
= - sin-1 -+- Va^^+K.
2 a 2
Example 7.— To find J ^^^
Let AT = a sinw i. e., u = sin-1 -
a
rf# ^(a sin u)
then -j— = -- T - - = a cosw
du du
also Vaz—x2 = a cosw, as before.
/" dx fa cosu.du __ .
J Va*=^~ J~a~^r-Jldu
= u+C
= sin-1 -+C.
a
Example 8.— To find /"-
X
In this case let AT = a sinh u, i. e., u = sinh"1 -
a
dx d , . . .
then j- = j- (a sinhw) = a coshM.
du du ^
Now cosh2 u— sinh2 u = i (cf. p. 291, Part I)
and thus cosh2 w = i +sinh2 w
or a2+Ar2 = a2 cosh2^
and Va2+#2 = a coshw.
[ dx fg coshu.du _ r -, , r
•• J Va*+^~* ~ •' a coshtT ~ Ju
= sinh-1 *-+C.
Referring to p. 298, Part I, we see that —
, . x . (x+ Vx*~d*}
cosh"1 - = logi -
a &l a J
and also sinh- * = I
~ - sinh-1 *+C
2 a
or
FURTHER METHODS OF INTEGRATION 151
r flx
Example 9. — Find the value of I - ._ and thence the value of
•' vx2— a2
f dx
J
Dealing with the first of these—
X
let x = a coshw, i. e., u = cosh"1 -
a
dx . ,
j— = a sinhw
du
or dx = a sinhw du
and x2— a2 = a2 cosh2w— a2 = a2 (cosh2w— i) = a2 sinh2**
fasinhudu
sinhw
f dx f dx fasin
J Vx2—az ~ J a sinhw J a si
= fdu
— u+C
— cosh"1 -+C
To evaluate the second integral, let x-\-a = X
then dx = ^X and J = f
Example 10. — Find the value of I 0-77 ~a
J Ct ~j X
The substitution in this case is —
a tanw for x
x
i. e., x — a tanw or w = tan-1 -
dx d . . 2
then j- = j- (a tanw) = a sec^w
du dx ^
and A-2+a2 = a2tan2w + a2 = a2 (i+tan%) = a2 sec2^
sec2w rfw i r,
= - /aw
/" ^ /"a s _
' ' J a?+x* ~ J a2 sec2w ~
= -M + C
a
= £ tan-1 - + C.
152 MATHEMATICS FOR ENGINEERS
f dx
By an extension of this result such an integral as I
J
may be evaluated; for let x-\-a = X
dX d(x+a)
then -s— = v ,r ;- = i
dx dx
i. e., rfX = dx
hence ( dx - f **X - ' tan-1 X4-C
J x + az + b* ~ J X^+F2 ~ b tc F^
i. e., I tan-1 ^~+C.
b b
The following examples are illustrative of algebraic substitution or
transformation.
r
Example n. — To find the value of I
Our plan in this case is so to arrange the integral that the method
of a previous example may be applied.
•—z — z—z—
Hence ( dx - f - -** __ = [ d*
J V2ax-xz J Va*-(x-a)2 J V«2-X2
the change from dx to rfX being legitimate, since , = -, — ' = i
ctx dx
and by Example 7, p. 150, the value of this integral is seen to be —
f
•
J
Example 12. — To find J
dx . x — a
= sin-1
xz a
dx
In this case the substitution is entirely algebraic.
i , , du i
Let u — then , = „
x dx x2
or dx = —x2du.
Then a*+x*S
dx f —xzdu f du
!\4 J I „•!,.•! I T\T vx ~3 J ^
[ udu
= ~r,
FURTHER METHODS OF INTEGRATION 153
To evaluate this integral we must introduce another substitution.
Let y = «2«2+i
then -^ = 2a2u
au
or udu = — -9dy.
zaz '
Hence the integral - ~ = ~~^ +C
+ C
__ I -4-0 = _ - __ I-C
~ 2 ~
L+c.
dz At
Example 13. — The equation -,. — . occurs in the statement
dt \/ j2 — j
of the mathematical theory" of fluid motion, which is of value in
connection with aeroplane design. Solve the equation for z.
To obtain z from -^ we must integrate with regard to t; and to
ctt
effect the integration let u — t2— i, so that -^ = , . — - = zt
du
or • dt — —r.
21
, Atdt fAtdu [Adu
I hen z =
2U*
or AVt2— i + C.
Many difficult integrals of the form / can be evaluated
J x(a+bxn)
by the substitution z — x~n.
For if z = x~n log z = — n log x
d log z _ d log x
j ^ 3 '
dz dz
i _ J. log x ^ dx
— — — ^ jj ^ ~3~
z dx dz
i i dx
154 MATHEMATICS FOR ENGINEERS
The integral f x(a+bx^ thus reduces to -!j(_, the value
of which is — log (az-\-b) or — log / , , A
na ' na ° \a-\-bxnJ
/dx
For x* write z~l, so that in comparison with the standard form
n = 7.
_i
~ 28 I0g
Example 15. — Find the value of / r.
J (i — 2x)*
It will be observed that the denominator is a surd quantity; and
in many such cases it is advisable to choose a substitution that
rationalises the denominator. Thus in this case let u* — i — 2X.
,_, QA& Ct/14/ GLIfi Ctr'VC
~dx = ~du X dx = 2UJ»
and -, (i — 2x) = — 2
du
so that 2Mj- = —2 or dx — —
dx
i—uz (i — uz\*
Also i — 2X = u-, whence — - = x and x* — (-- j .
2 \ 2 /
Expanding by the Binomial Theorem —
x* — ~ (i — 4«2+6M4— 4«6 + w8).
/" A;*^
Hence
J (l-2^)i
- _L /"(i — 4M2+6M4 — 4M6+M8)X — udu
~ i6J u
i ( 4U3 6u5 ^u1 . u*
= ~^(u- T+T 7 +
4tf,^_i-
3 5 7
16
which result could be further simplified if desired.
FURTHER METHODS OF INTEGRATION
155
The next example introduces the substitution of an algebraic
for a trigonometric function.
/dx
-. —
sm x
Since sin 2 A = 2 sin A cos A, then sin x = 2 sin - cos - .
2 2
f dx if
Hence — — : = -
/ sm x 2 / .
J J Cl
Now let
then
or
dx i
dx
. x
. X XI
sm
2 .1 -'•
sin - cos -
— X COS2 -
2 2
X 2
COS -
,
2
9 #J
sec2 a*
I
2
~~ 2
X
tan -
*7
u = tan
2
du i „ x
-— = sec2
dx 2 2
sec2 -
2
sn x 2 x
sec2 .
= J|' = log «+C
= log tan -+C.
Integration by Parts. — When differentiating a product, use
is made of the rule —
d i x du . dv
dx(u^=Vdx+Udx {Refer p. 70.!
If this equation be integrated throughout, with respect to x —
uv = fvdu+fudv
or, transposing —
fudv = uv—fvdu.
Many products may be integrated by the use of this rule.
156 MATHEMATICS FOR ENGINEERS
Example 17. — To find f^x.exdx.
Let « = 4#, i- e-, du = ^dx
and let dv = exdx, i. e., v = ex.
Then f^x.exdx = fudv
— uv—fvdu
Example 18.— Find f^.e^dx.
Let u = 5#2, i. e., du = loxdx
and dv — e*xdx, i. e., v — -e4*.
4
Then f$x2.e*xdx = fudv
= uv—fvdu
= sx* . Te*x— r
4 M
= ^x2.e*x—5 I xeixdx
4 -2.1
^-r r A-TJ i n f1 in rwhere u = x -i
Now fxe*xdx = x.-etx— e*xdx
4 J 4 [ and v - VJ
= ^.e4a;-4e4a:
4 16
+ C
Example 19. — To find Te0* sin (bx+c)dx and also —
fe^ cos (&Ar+c)^Ar.
[The two integrals must be worked together.]
Dealing with the first, which we shall denote by M —
Let u = sin (bx-\-c), then du — b cos (bx-\-c}dx
and dv — e^dx, so that v — fe^dx = e0*.
Then M = - e™ sin (6Ar+c) — - eax b cos (&#+c)d*
flS ^ &
= - e^sin (6^+c) — fe™ cos (bx+c}dx
N ......... (i)
(I (t
where N stands for the second integral whose value we are finding.
FURTHER METHODS OF INTEGRATION 157
By developing this second integral along similar lines we arrive at
the value —
N = -ea:rcos (bx+c) + -M. ...... (2)
We have thus a pair of simultaneous equations to solve.
Multiplying (i) by b and (2) by a and transposing —
b bz
6M = - eax sin (bx+c) -- N
a 'a
feM = — eax cos (&#-|-c)+aN.
Subtracting o = eax f~- sin (bx+c) -\-cos (bx+c) \— N(-4-a)
\__Qt — J \ fl£ '
t, -NT nrrb sm (&#+c)+a cos
whence N == eax\ '
- — z ,a —
•u jit. .• HT /r^R* sin (&#+c)— 6 cos (fof+c)~i
and, by substitution, M = eax\ a8+68 —
y^ao; sin (bx-{-c)dx = 2 ,3 [a sin (bx-\-c)— b cos
~
and ye^ cos (bx+c)dx = 2 a [6 sin (6^+c)+a cos
Example 20. — An electric current i whose value at any time t is
given by the relation i = I sin pt is passed through the two coils of
a wattmeter; the resistances of the two coils being Rx and R2
respectively, and their respective inductances Lx and L2. Then to
find the separate currents in the two branches it is necessary to
evaluate the integral —
fQe*'dt where P =
and Q = J-^TT~ sin pt+-. — —^ cos
Evaluate this integral.
j j
Q = f rT~(Ri sm pt + pLi COS pt) = = —
where c = tan-1 ^~ (see Part I, p. 276)
**
or Q = M sin (^>^+c), where M = T-^
Then fQe™dt = fe^M sin (pt+c)dt = Mfe™ sin (pt+c)dt
and this integral is of the type just discussed; its value being —
[P sin (pt+c}—p cos
and in this form it is convenient to leave it, since in any numerical
application it would be an easy matter to evaluate P, M and c before
substituting into this result.
158
MATHEMATICS FOR ENGINEERS
Some miscellaneous examples now follow, involving the use of
the methods of this chapter.
Example 21. — Find the value of
Let
». e., 5 =
Let x = — 5, then 5 = o — 2B
B = -2-5.
Let # — — 3, then 5 =
^_o._B
x+5
*+3).
A = 2-5, i. e.,
= 2-5 2-5
p
J j x
5 ^#
*2+8*+i5 x+3 x+5
= ft 2-5 dx f* 2-5 dx
= 2-5. flog (^+3)— log
= 2-5 [log 5— log 7— log 4+log 6]
= 2-5 [1-6094 — 1-9459— 1-3863 + 17918]
= 2-5 X -069 = '1724.
As an alternative method of solution, the graphic process of
integration possesses certain advantages in a case such as this.
It might even be advisable, in all cases of definite integrals
where the algebraic integration involves rather difficult rules, to
treat the question both algebraically and graphically, the latter
method serving as a very good check on the accuracy of the
former.
f2
In this example /
where
5 x -
V =
/
J i
=51 ydx
-'
hence it is necessary to plot the curve y = -. . . . and find the
(X~r3)(x~r5)
area between it, the axis of x and the ordinates through x = i and
x = 2.
The table for the plotting reads —
X
i
1-2
1-4
1-6
1-8
2
y
•04167
•03841
•0355
•03294
•03064
•02857
and from these values the curve AB is plotted in Fig. 37.
FURTHER METHODS OF INTEGRATION
159
The sum curve for AB is the curve CD, the last ordinate of which,
measured according to the scale of area, is -0095. This figure is the
area between the curve AB, the ordinates through x = i and x = 2,
and the base line through y = -025; and hence the full area under
the curve AB = -oo95 + area of a rectangle -025 by I, i. e., -0095 + -025
or -0345.
dx
= '°345
Thus
= 5 X -0345 - -1725.
040
y
035
03O
0£5
D
Sum Curve
polar distance— '8
•010
•OO95
•008
•OO6
OO4
•002
o
FIG. 37. — Graphic Integration of Ex. 21.
dx
Example 22. — Find the value of I — 2 .
XT
Now
f dx i . _ (x— a) , „
I -5 , = — log C 7 — — ( (see Example i, p. 147)
/ #2— a2 20 (x-\-a) '
cM)
i i , V • 2/
rf* i r ^
T r^r a^
A i *-^l *"& *1 /
= — log V
12 &
or
(ix+3)
160 MATHEMATICS FOR ENGINEERS
Example 23. — Find the value of I —4 ----- , — dx.
J 4 gx2— #+i2
gx
In this case the numerator is of the first degree in x, whilst the
denominator is of the second degree. Also we notice that the
derivative of the denominator is i8# — 5, and the numerator is
4(18^—5). Thus the derivative of the denominator and the numerator
are alike except as regards the constant factor 4. Hence the substitution
will be u for gx2—
If u = 9#2— 5*4-12, -r- = i8#— 5 or du = (i8x—$)dx
(IX
so that (j2X—2o)dx = 4(i8x—5)dx = ^du
f 72* -20
J tx2 —
:=4 U
Example 24. — To find the value of I
This is evidently of the type I
= 4 (loge 418— loge 136)
= 4 (loge 4-18— loge 1-36)
= 4 (I-4303 — 3075)
= 4-49-
dx
dx
(^4
for
so that a = 3 and b = V&.
r dx i.i x-\-a _ . , _
— = r tan-1 — ± — \-C (cf. Example 10. p. 151)
J xz+6x+i$ b b
Example 25. — For a single straight wire at a potential different
from that of the earth, if v — radius of wire in cms., / = length of
wire in cms., a — surface density of charge in electrostatic units per
sq. cm., then the potential P at any point on the axis of the wire due
to the charge on a length 8x is given by —
ft dx
so that the potential at the middle point = mrffj l , z-
Evaluate this integral.
FURTHER METHODS OF INTEGRATION 161
/dx
, „
Va2+#2
;. f^== = log(*+ V*2+r*\ (cf. Example 8, p. 150)
Pm = 27WO-
r (' + J*:.
i — ir-
= 27Wo- log-
\/5+*-;
or TT^O- log-
* where d is the diam. of wire.
The following example involves the use of three of the methods
of this chapter.
Example 26. — Find the value of /Vs— ^-- L — ; — i-
I ( x \~ 'ZX ~\~ 1 1 \X~\~ ^ )
The fraction under the integral sign should first be resolved into
partial fractions.
t. e.,
Let x = — 3, then —
— ii = C(9— 6+7) = loC
i. e..
££
10*
Values of A and B can be found by equating coefficients of x and
also those of x2.
By equating coefficients of xz o = A+C and hence A = -
162 MATHEMATICS FOR ENGINEERS
oo 22
By equating coefficients of x 5 = 3A + B+2C = _- + B— — .
10 10
10
II
*+3
i f n*+39 i
Hence the fraction-
We can make the numerator of the first of these fractions into
some multiple of the derivative of the denominator; thus —
The derivative of the denominator —
and the numerator =
and if u —
then du = 2(x+i)dx
and
= du-\-28dx
frr.
J (xi-
loJ xz+2x+j x+3)
iidx
i / [(iix+3Q)dx fi
IO\J Xz + 2X+7 f X
i_(fii, f 28dx
~~ 1O\J 2U J
28 ^4-1^ \ rx2+2X+?
/
V6
- g log (x* + 2x + 7) + 6 tan- -~ log
or log (*!±?*±7).** .14
\ rx2X? -,
=^2+2^+i + 6
} L^^I22j
V6
Example 27. — An integral required in the discussion of probability
/-co
is / e~x*dx. Find a value for this.
J o
/GO
e~x*dx.
Replace x by ax and thus dx by adx.
Then I = I e-^&adx.
J o
FURTHER METHODS OF INTEGRATION 163
Multiply all through by e~at.
f
Then Ie~at = I e-**()-+&').adx and integrate throughout with
regard to a; thus —
/<*> rx=x> ra=<x>
le-a'da = e-a^+x^adadx.
0 J x=~Q J o=0
i~<*> rco r«> r
But / Ie-a'da = 1 1 er^da = 1x1 since / e~&dx and /
J 0 ^0 J 0 •'
have the same value,
hence I2 = /*" ' ¥*™* e-+
.' x=Q J o=0
(i)
The value of the double integral on the right-hand side will be
found by integrating first with regard to a and then with regard to x.
Dealing with the " inner " integral —
/<j=oo
a=0
let A = a2, then dA. = zada, and let M represent
Then P~™ e-<Kl+'V . ada = f
J 0=0 J 0
o
2M\ /o o and oo also
i / _AM\°° since the limits for A are
2M /•
_ _
~2M."> ~ 2M-
Referring to equation (i) and substituting this value therein —
1 / \°°
= - /tan"1 x ) (cf. Example 10, p. 151)
= - (tan-1 oo —t
2 v
I/7T \ 7C
= -I -- O) = -.
2\2 } 4
= - (tan-1 oo —tan-1 o)
2 v
As an extension of this result it could be proved that-
Reduction Formulae. — Many of the exceedingly difficult
integrals which arise in advanced problems of thermodynamics,
164 MATHEMATICS FOR ENGINEERS
theory of stresses, and electricity may be made by suitable sub-
stitutions to depend upon standard results obtained by a process
of reduction. To grasp thoroughly the underlying principles on
which the process is based, it is well to commence with a study
of the simpler types.
7T 7T
[z /"a
We desire to evaluate the integrals I sin"0 dQ, I cos"0 dQ
-' o -'o
and I sinm0 cosn0 dQ, where m and n have any positive integral
J n
0
values.
Taking the case in which n = m = o, we have the results
IT
fz
reducing to the form / idQ, the value of which we know to
J
. 7T
be -
2
If m = n = i
ir w
/"2 / \ 2
I sin 0^0 = — (cos0) = —(cos 90°— cos o°) = I . . (2)
Jo /o
IT If
IZ cos 0 dQ = ( sin 0Y = (sin 90°— sin o°) = i ... (3)
J o /o
from which pair of results we may say —
7T 7T 7T
/ 2 sin 0 dQ = r cos 0 dQ = ] sin ( ?r— 0)^9
/ / / \ 2 /
or more generally —
jaf(x)dx = j*f(a-X)dX (4)
result of great usefulness.
7T W If
Also \ sin 0 cos 0 ^0 = - \ sin 20^0 = — I cos 20 )
./o 2./o 2X2\ /o
i , ,
= (COS 7T— COS O)
4
=J .... (5)
By the process of reduction of powers we may express the
integral to be evaluated in such a way that it depends on results
(i). (2), (3) or (5).
FURTHER METHODS OF INTEGRATION 165
Thus—
77 tr «; 77
f2 sin26 dQ=* I* (i-cos 20)^0 = - I (\dQ- f* cos 20 do]
J o 2 J o 2 \_J 0 y0
I/IP N
= - ( O
2 \2 /
cos28 rf0 = 2 sin20 rf0 = -.
o Jo 4
and from equation (4) —
77
;
f 0
ft
Now let n = 3, i. e., we wish to evaluate I sin30 dQ.
J o
It 77 77
Then ] sin30 dQ = \ sin20 . sin 0^0 = ] (i— cos20) sin 0 dQ
J o Jo Jo
IT
= — I 2 (I— M2) rf«
•'o
(o and ^ being the limits for 0 J
u being written for cos 0 and — du for sin 0 dQ, since — -^ — = —sin 0
and thus -^ = — sin 0.
77 17 „.
, T f "~ ^ / W3\ ~ 2 / COS30 \2
Now I (i— w2) aw = I M- — = cos 0 )
Jo = o v V 3/e = o \ 3/0
Thus f2 sin30 dQ = — I * (i— u2) du = +-
^o /»-o 3
and if n be written for 3 we note that the result may be expressed
77 77
in the form I sinM0 ^0 = and also I cosw0 dQ = -
J o n Jo
n being an odd integer.
77
fz
Let n = 4, then I sin40 ^0 is required.
J o
Now sin40 dQ = sin30 . sin QdQ = udv, where u = sin30,
and dv = sin 0 dQ or v = —cos 0.
166 MATHEMATICS FOR ENGINEERS
7T 7T 7T
Hence ] sin40 dQ = —(cos 6 sin30Y— 1 2— cos 0 . <Z sin30
J 0 / 0 J 0
7T 77
/ \ 2 /*2
= — ( cos 0 sin30 ) + I cos 8 . 3 sin20 cos 0 dQ
\ /o J o
7T
= 0+3 1 sin20 cos20 dQ
J o
77
/ \2
since ( cos 0 sin30 ) = (o X i) — (i X o) = o.
'o
7T 7T
Now I sin20 cos20 ^0 = ] sin20(i— sin20)^0
Jo J o
= (sin20-sin40)<f0
o
It 77 IT
Hence— \ sin40 dQ = 3 ] sin20 dQ—3 (2sin40 dQ
J o J o J o
[I [I
or 4 sin40 ^0 = 3 1 sin20 dQ
J o Jo
7T 7T
f 2 q /"§
and I sin40 dQ = ^ sin20 dQ.
J o 4J o
We have thus reduced the power by 2, and knowing the result
7T If
for / sin20 dQ, we can finally state the value for / sin40 dQ.
Jo Jo
7T
TVinc cin^fl /7A — _ V _ — ^ /^r- — V — S/ _
lllUo oill U Ct-U ^— /\ -^ x- Ul x\ /\
J0 4 4 16 422
fi (n— i) («— ^ «•
or I sin 0 ^0 = '- X, *» - , n being an even integer.
J Q W (W — 2j 2
In like manner it could be shown that —
7T 77
/•<j tf T /•§
sin50 ^0 = °— - / sin30 dQ.
Jo 5 J o
Thus sin50^0 = ^x" or ~
5 5-2 5-3
FURTHER METHODS OF INTEGRATION
IT
which is of the form f2 sin"0^6 = ^^ . ^
J0 n n—2
n being an odd integer.
Similarly —
JT It
• «0 JO 6 — lfZ . 6—1 6 — 3 6 — 5 TT
sm66 dQ = —£- I sin4^ = — ^— X z — - X £ — - X -
6 ;0 6 6—2 6—4 2
5.3.1 TT
or B-^ — x -
6.4.2 2
which is of the form —
OBfl A V H* V . . y .
o n(n— 2)(n— 4) 2
w being an «;<?« integer.
Summarising our results —
sn = cos =
^o n
if « is an evew integer.
P sinw0 ^6 = P coS*0 dQ = etza^K-nSj^^a
70 Jo n(n—2)(n—4) . . . I
if « is an o^ integer.
JT
/2
sin9 ^</^.
o
In this case w = 9 and is odd.
Hence — sin90 d9 =
/•• .
su
J o
8.6.4.2 _ 128
9.7-5-3 315
w
fz
Example 29.— Evaluate / cos100 dd.
J o
Here n = 10 and is even.
Hence
Jo
. .
10 .8.6. 4. 2 2 512
Example 30. — The expression — I ^3 — dt gives the theo-
7C-|~4-' °° *
retical thrust on a plane moving through air. Evaluate this.
V/«=i = *tzi-~ = tVi-sin2u = t
if sin u is written in place of -.
168 MATHEMATICS FOR ENGINEERS
i d sin u dtr-1 dt~l dt
Then since sm u = - — t~l — ^ = -3— = -rr X T-
or cos u = —-5 j-
t2 du
du cos u
whence dt = =— ^ — .
Again, when t = x> - = o i. e., sin u = o or u = o
t
and when t = i 7=1 i- &•> sin u = i or u = -.
t 2
Hence —
TT It
f1 Vt2—!,, f2 cos u sin'u cos udu f2 , TT
— is — <# = — I = r- = — = — / cos2waw = .
J ao t3 Jo sinttsm2w Jo 4
Thus -^ / '-^— -«8 = -^F- X -- = -^-.
4 ^ ~r~4
We can now direct our attention to the determination of the
r\
value of I sinm0 cos"0 dft, where m and n are positive integers.
' o
It is convenient to discuss a simple case first, viz. —
TT
[z
sin20 cos 0 dQ.
•' o
T , . on , , du d sin20 d sin 0
Let u = sm20 so that -^ = 5—^-5 X — ^~
^0 rf sin 0 <?0
= 2 sin 0 cos 0
and du = 2 sin 0 cos 0 ^0
also let dv = cos 0 <£0 so that v = sin 0.
Then, by integrating by parts —
7T IT 7T
f 2 sin20 cos 0 ^0 = (sin3 0V— f 2 2 sin20 cos 0 ^0
J o \ /o J o
/2 / \2
sin20 cos 0 ^0 = ( sin30 ) = i
o /o
•n
[2 I
or sin20 cos 0 ^0 = -
^o 3
j . , , , ...
and might be written as
.
m+n
FURTHER METHODS OF INTEGRATION
169
Example 31. — H, the horizontal thrust on a circular arched rib
carrying a uniformly distributed load w per foot run of the arch, is
obtained from —
(X-si
V*
sin2 e (cos 0--S66) dQ
2R3 I6 (cos 0 — 866) dQ
J o
if the span is equal to the radius of curvature (see Fig. 38).
If w = -5 ton per foot, and the span = 60 ft., find the value of H.
Here w = -5, R = span = 60.
IT
[8 (J-sin2 6) (cos 6- -866) dQ
Hence
-5X60
(cos 0- -866) <Z0.
FIG. 38. — Circular Arched Rib.
Dealing with the numerator separately, as this alone presents any
difficulty —
(^— sin20) (cos 0— -866) = - cos 0— -2165— sin20 cos0+-866 sin20.
\4 4
Then [(-— sin20] (cos 0 - -866) dQ
= l~ cos 0d0 -f-2 165 d0-/sin20 cos 0 dQ+f-866 sin20 dQ
but, as proved on p. 148, /sin20 dQ = ~fidQ—-fcos2Q dQ
= sin 20
2 4
and as proved on p. i68ysin20 cos 0 dQ — - sin30
170 MATHEMATICS FOR ENGINEERS
w
thus r (i-sin20) (cos 0--S66) dQ
J o H
ff TT n- TT
= (-sin 0)°~- (-21650)°-* (sin30)V433 («-^ sin 28)*
-o)-f (<.S).-o)+.433 (Hx -866-0+0)
= -125— -1134— -04I7+-2267— -1875 == -0091.
Dealing now with the denominator —
IT n
(* (cos 8— -866) dQ = (sin 0- -8660Y = -5- -866 X ~ = -5 — 4534 = -0466.
J o '•
TT 15 X -0091
Hence H = -* - ^— = 2-93 tons.
•0466 yj
Carrying the investigation a step further, let us discuss the
case of I sin20 cos20 dQ.
Jo.
»T It IT IT
f 2 sin20 cos20 dQ = I * (i— cos20) cos20 dQ = f2 cos20 dQ— 1 2cos40 dQ
Jo J o Jo J o
and from the previous result this value = ™— ^— -
4 4.22
_ I 7T I I 7T
~82 4*2'' 2
This result might be regarded as obtained by first reducing the
power m by 2, and next that of cos20 by 2.
Thus for the first step —
7T 7T
f 2 sin20 cos20 rf0 = ^^ f 2
Jo 4 70
cos20 ^0
and for the second step —
IT
idQ
f 2 cos20 ^0 = ^^ f 2
Jo 2 J0
2—1 TT n—i TT
- -,-r. _ /"\T*
• \JL
2 ' 2 W+w 2
m, for the second integral being zero.
FURTHER METHODS OF INTEGRATION 171
rr
ft
In a similar fashion we might reduce I sin40cos30^0 as
•' o
follows : —
First reduce m by 2, then —
1C IT
! * sin46 cos30 ^0 = 4-^ (* sin20 cos30 dQ.
J o
4+3 J o
Next reduce the power of sin20 again by 2.
7T T
Thus P sin40 cos30 dQ = £f* . *^ I *
Jo 4+3 2+3 J0
cos30 dQ.
Now reduce the power of cos30 by 2, and remember that m
is now = o. Then —
sin'e cos'erfe _ 1=5 . ?=I . 3=1 f*
o 4+3 2+3 3+o^o
.
7-5-3
/2
In the evaluation of the integral I sui"'0cosn0^0 we thus
* o
proceed by reducing by 2 the powers of m and n in turn until they
become i or o. The various cases that arise are —
(a) m and n both even : in which case the final integral is
f * rf0 = ?
;0 2
(6) w and w both odd : in which case the final integral is
IT
[2 !
sin 0 cos 0 dQ = -.
J o 2
(c) m even and n odd or vice-versa : in which case the final
•a it
fz [2
integral is either / cos 0^0 or I sin0^0, the value of either
J o J o
being i.
The results for the three cases can be thus stated —
(a) m and n both even —
Psi
J0
sinm0
o
(n-i)(«-3)x . . . i *
n(n— 2)x ... 2 2
I72 MATHEMATICS FOR ENGINEERS
(b) m and n both odd — •
. (m_I)x(w— 3)x ... 2
smw8 cosn6 <ft = / — , v . . ' — *— : •:' — ? — ; — r
(m+») X (m— 2+n) X ... (»+3)
(tt-i) x(n-3) ... 2 i
X(«+I)X(«-I)X . . . 4X2
since, after reducing the power of sinm0 by 2 at a time, we must
be left with sin 0 cosn0, so that the value of m to be used in the
reduction of cosM0 must be taken as i.
(c) m even and n odd
fl l«
I smw0 cosre0 ^0 =
.' n
(m+n) X (m—2+ri) X ... (w+2)
»-ix«-3)X . . . 2
X
7T
ft
Example 32. — Evaluate I sin?0 cos100 rf0.
/ 0
This is case (a), i. e., with both m and w even.
7T
Hence 2 sin»0 cos"0 dQ = - -' - -- -'^- X
f 2 si
/0
- - -- -
.10.14.12 10.0.6.4.2 2
1179648'
7T
[z
Example 33. — Find the value of I sin30 cos50d0.
J 0
This is case (6), i. e., with both m and n odd.
7T
Hence / 2 sin30 cos50 dQ = \ X J— X - = - .
Jo 8 6.4 2 24
IT
Example 34. — Find the value of / 2 sin70 cos*0 ^0.
/ 0
This is case (c), but with m odd and n even.
it
Hence T sin70 cos40 d0 = .^' j'^ X ^ X i = l6
' 0
n-9-7 5-3 ii55_
The value of the foregoing formulae is found in their employ-
FURTHER METHODS OF INTEGRATION 173
ment in the evaluation of difficult algebraic functions, which may
often be transformed by suitable trigonometric substitution.
Thus to evaluate / ^-^(i—x^^dx, known as the First Eu-
J o
lerian Integral and usually denoted by the form B(m, n), we may
substitute sin20 for x. Then, since when x = o sin20 must = o
and thus 0 = o, and when x = i sin2 0 must = i and thus
/
xm-1(i—x)n-1dx = sin2n*-20(i— si
~dx d sin2 6
_ d sin2 6 d sin 6
~ d sin 6 ' ^0
_ = 2 sin 0 cos 0
= ] sin2*-2 0 cos2n~2 0 x 2 sin 0 cos 0^0
-'o
ir
fz
J n m
and this can readily be evaluated.
Example 35. — Evaluate / x*(i— xz)?dx.
J o
Let x = sin 0 then i— xz = i— sin20 = cos20
dx d sin 0 ft
,
and
Also when x = o sin 0 = o and thus 0 = o
and when x = i sin 0 = i „ „ 0 = -.
Then —
v 5
I *4(i-xz)*dx = f 2 sin40 cos50 dQ cos 0 = f 2 sin40 cos60 dQ
-' 0 J 0 .' 0
10. 86. 4. 22
Another important result obtained by the process of reduction
r*
is the value of I e~xxndx. This is termed a Gamftia Function;
J o
this particular integral being the («+i) Gamma Function denoted
Thus f e~*x»dx =
J o
/-co
and I e-t&^-dx = Y(n)
J o
the latter integral being also called the Second Eulerian Integral.
174 MATHEMATICS FOR ENGINEERS
To evaluate r(w+i) let u = xn and dv = e~xdx
so that v = —e~x and du = nxn~ldx.
,00 XOO /•»
Then I e~xxndx = ( —e~xxn) —I —e~x.nxn~ldx
Jo \ /o J o
xoo /•«.
= ( — erxxn ) 4-n\ e~xxn~ldx.
\ /o J o
Now when # = oo e"^" = — ^— , which can be proved = o
£ •
I X O
and when x = o e~xxn = - - = o.
e
/•oo /•»
Hence / e~xxndx = n I e~xxn~1dx
Jo Jo
or r(«-{-i) = nT(n).
In like manner it could be shown that —
T(n) = (n—i)T(n—i), and so on.
If n is an integer it will be seen that we finally reduce to r(i),
f°
i. e., I e~xdx, the value of which is— (e~y> —e°) = — (o— i) = i.
Hence r(«-fi) = n(n—i)(n—2) . . . i = |_«.
This last relation will not hold, however, if n is not an integer,
but the general method of attack holds good; and tables have
been compiled giving the values of T(n) for many values of n,
whether n be an integer or fractional. Thus if an integral can be
reduced to a Gamma function or a combination of Gamma func-
tions, its complete evaluation may be effected by reference to the
tables.
/oo. _a^
e~h'dx by the
aid of the Gamma function.
Let X - ^ then ^ = -f- = ±*
«2 a# a# h2
, ,Y _ 2^^r
~ "P~
A2 rfX A2 dK. hdX.
or dx = = — ; — 7=^ =
2 x 2 h\/x. 2\/X'
Then f.TW, = /"e-xx^|- = *r.-*X-ta = *xr(?)
^0 •/ 0 2 VX 2-' 0 2 \2J
and the value of T( \ *s VTT^
FURTHER METHODS OF INTEGRATION
r «> _a* fr _
Hence I „ e h*dx = - Vie.
JO 2
175
On comparing with Example 27, p. 162, where a rather simpler
form of the integral is evaluated, we see the great saving effected
by the use of the Gamma function.
LIST OF INTEGRALS LIKELY TO BE OF SERVICE
f(ax»+b) to
fae^dx
fba^dx
fa cos (bx+d) dx
fa sin (bx+d) dx
— — .
n log a
= sin (bx+d)+C.
= — ~ cos (bx+d)+C.
fa tan (bx-\-d) dx = —, log cos (bx-\-d)-{-C.
fa cot (bx+d) dx = ^ log sin (bx+d)+C.
= | tan (6*+<Q+C.
= — cot (bx+d)+C.
fa sec2 (bx+d) dx
/a cosec2 (6*-H)
fa cosh (6^+^) to
/a sinh (bx+d) dx = ~ cosh (bx+d)+C
= T sinh
fa sech2 (6^+^) to = | tanh
/a cosech2 (bx+d) dx = — | coth (bx+d)+C.
* sin (6^+^) to = -
f"c0s (bx+d) dx *=
sn
sn
— cos
cos
f ax
\ +/~2 '
J V U X
(-T
JVx
dx
dx
176 -MATHEMATICS FOR ENGINEERS
r dx i ( bx+d\ . n
I • ~~'/i~ ~r~j\ == i 1°S ( tan +C.
J sin (bx+d) b ° \ 2 /
f dx i
]0x+b =-log(**+fc)+C.
/tan (bx-}-d) dx = — -? log cos (bx+d)+C.
/cot (bx-{-d) dx = r log sin (bx+d)+C.
r dx i , , x . r
I -o-j — « = - tan * -+C.
J «2+A;2 « a
dx 1,1 ^+« , /-
= *tan T+c-
r-r^i = X log^+C.
J a2— xz 2a
f dx i_ ,
J (x-\-a)z — bz ~ 2b °
dx
= sm~
1-7===, = log ^~rv-* -•* j+c.
J v x a
r dx
}'•
r (ax-\-b)ax i . ,
J ^tf^bx+d) ~ 2 °g (a
Tcosec (ax-\-b) dx = - log (tan ' -)+C.
** d \ 2 /
/ sec (a^+&) <fo = - log ta
/^ i , x . „
. — = - sec"1 -+C.
x\xz — uz & a
f dx , x . r , x r
I . = vers"1 -+C or i— cos x -+C.
J V2ax—xz « a
l—xzdx = -y«2— xz+2 sm
__ x ez
jVxz-az dx = ~Vxz-az- •
FURTHER METHODS OF INTEGRATION 177
fV(x+a)*-Pdx = I(x+a)V(x+a)*-b*-~ cosh-1 *±^
*2
-a? dx = ~Vx24-a24-— sinbr1 -
2 2 a
/V/(*+a)M-6* <& =
^ -5
/" sin2 ;*; <fa = sin 2^+C.
2 4
. A; I
/ cos2 x dx = -4— sin 2x-\-C.
2 4
/* sinm x cosn A; ^ = — /" sinm~2 x . cos" ^ dx
J m-\-nJ
m-\-n
[ dx
J (aTI^ji
I ^-*i^
* o
/•oo «*
I e^w dx
Jo 2
tr T
sin20 ^0 = | 2 cos20^ 0 = "".
J o 4
IT
-7
sinm-l ^ , cOSn+1 # r
f-L/.
sin«6 & = cos«6 <*0 = ~
0 »(»— 2)(»— 4) ... 2 2
if w is an even integer.
sin«6 ^0 = cos«0 ^0 = *n7- • . . 2
fsi
Jo
.
o »(«— 2)(»— 4) . . . I
if w is an odd integer.
(«-l)(»-3) . . . I 7T
X
IT
r1 •
I SI
J A
if w and w are both even.
sinw0 cosw0 dQ= y\ /^
0 (w+w)(w+w— 2) . . . (w+3) (w+i)(w— i) ... 4 2
if w and n are both odd.
178 MATHEMATICS FOR ENGINEERS
It
(*
I sin'"
' o
e cos»e rfe = -= x = ' ' '
—2) . . . (n-\-2) »(»— 2) . . .
i —
if m is even and n is odd.
Pdv (where pvn = C) = (v^~n — v^
I pdv (where pv = C) = C loge -2.
= a log p
sin"1 a#*fo = x sin~1a#+;_ Vi — a2xz-\-C.
tan-1 ax dx = x tan"1 ax log
20, b
f (a2-^2)1 ^ = -(a2-*2) §+
J 4
§3 - sn
Exercises 15. — On Further Integration.
Evaluate the integrals in Nos. i to 18.
dx_ 9 f dx f (x-i)dx
*—*' J *x*+6x+2i' 'Jc>x2-i8;
4. - | VJ 3 , which occurred when finding stresses in a crane
hook.
5. t — 95-5 1 (6—h}*dh, referring to the time for emptying a tank.
7T
2 /"2
6. ^3~ W2dy. 7- sin 5^ cos '
ft ^ 4 A /"sWR/i i . rtN-, _ ,ft
9. ^^« — . 10> / -Frrl --- sin 6 R2cos6d6.
sin2 5^ -' o El \TT 2
11. /4 tan 5/ <tt. 12. /sin-1 ^r dx. 13.
7T
14. /5e3:c sin 2Ar rf^r. 15. J ,^. *2.2. 16. J 5^ sin ^r ^.
17. /cos 6 sin2 6rf0. 18. h = f [-*-&* ,j~,rr&, relating to the
J o 2gn2 (d+kx)5
flow of water through a pipe of uniformly varying diameter; the
diameter at distance x from the small end being = small end diam.
-f-**.
FURTHER METHODS OF INTEGRATION 179
19. The time taken (t sees.) to lower the level of the liquid in a
certain vessel having two orifices in one side can be found from
nodh
at =
•12 + Vh
Find this time if the limits of h are o and 10.
20. Express e~l* as a series, and thence find the value of —
2 (* »
VrcJ o
21. The maximum intensity of shearing stress over a circular
F rr i
section of radius r = S = — j. / 2(r2— yz)*ydy.
If I = — , find a simple expression for S.
22. Evaluate the integral I - _____ [Hint. — Rationalise the
J t— vr— 4
denominator.]
«•
23. Write down the value of |2 cos8 QdQ.
J o
24. If the value of log r(i-85) is given in the tables as 1-9757,
ra>
find the value of I e~xxl'**dx.
-' o
w
25. Write down the value of 1 2 sin2 6 cos7 0^0.
J o
26. When finding the forces in a circular arched rib it was found
necessary to find the value of —
ir
2j6R3(cos2 6— Vs cos 6+^)^0. Evaluate this integral.
27. Evaluate the indefinite integral fVzi — y(— xz dx.
28. The attraction F of a thin circular disc of radius r on a body
on the axis of the disc, and distant z from the centre, is given by —
rdr
-
where a is the density of the disc, and k is a constant.
Find the value of F for this case.
CHAPTER VII
MEAN VALUES: ROOT MEAN SQUARE VALUES:
VOLUMES : LENGTH OF ARC : AREA OF SURFACE
OF SOLID OF REVOLUTION : CENTROID : MOMENT
OF INERTIA
Determination of Mean Values. — It is frequently necessary
to calculate the mean value of a varying quantity : thus if a
variable force acts against a resistance, the work done will be
dependent on the mean value of the force; or to take an illustra-
tion from electrical theory, if we can find the values of the current
and electro-motive force at various instants during the passage of
the current, then the mean rate of working is the mean value of
their product.
The mean value of a series of values is found by adding the
values together and then dividing by the number of values taken.
If, however, a curve is drawn to give by its ordinate the magnitude
of the quantity at any instant, the mean value of the quantity is
determined by the mean height of the diagram, which is the area
divided by the length of the base. This really amounts to the
taking of an exceedingly large number of ordinates and then
calculating their average.
The area may be measured by the planimeter, in which case
the instrument may be set to record the mean height directly, or
by any of the methods enumerated in Part I, Chap. VII.
A clear conception of the idea of mean values can be gained
by consideration of the examples that follow; the first example
being merely of an arithmetical nature.
Example i. — The corresponding values of an electric current and
the E.M.F. producing it are as in the table : —
c
o
1-8
3'5
5
6-1
6-8
6-1
5
3'5
1-8
E
0
9
17-5
25
30'5
34
30-5
25
17-5
9
1 80
MEAN VALUES
181
Find the mean value of the power over the period during which
these values were measured.
current x E.M.F., and
61-25
16-2
The power is measured by the product
thus the values of the power are —
o 16-2 61-25 125 186-05 231-2 186-05 I25
the sum = 1008-2 the number of values = 10
1008-2
hence the average or mean value = = 100-8.
A better result would probably be obtained if the values of the
power were plotted and the area of the diagram found.
Thus in Fig, 39 the base is taken as 10 units (merely for con-
venience), and the area is found to be 1013 sq. units. Then the mean
height of the diagram, which is = 101-3, is also the mean value
of the power; and a line drawn at a height of 101-3 units divides the
figure in such a way that the area B is equal to the areas A+A.
eoo
Rawer
-•""Mean hcighf
Ol E3456789IO
FIG. 39.
Let it next be required to find the mean value of a function —
say, the mean value of y when y = 4xz-\-jx—$, the range of x
being i to 6. We have seen that it is really necessary first to
find the area under the curve y = 4#2+7#— 5 within the proper
boundaries and then to divide by the length of the base; and
since the relation between y and x is stated, it is possible to
dispense with the graph and work entirely by algebraic integra-
tion; thus also ensuring the true result, for in reality the mean
is automatically taken of an infinite number of ordinates.
In the case taken as an illustration, the base is the axis of x,
or, more strictly, the portion of it between x = i and x = 6, so
that the length of the base = 6—1 = 5 units, and the area
between the curve, the axis of x and the ordinates through x = i
f6 f6
and x = 6 is given by the value of I y dx, i. e., I (4xz-\-jx— 5) dx,
182 MATHEMATICS FOR ENGINEERS
so that the mean value (for which we shall write m.v.) is —
*2+*_ dx
m.v. =
_[3+2J"~5*]'i
5
It is instructive to compare this result with the results obtained
by the use of the mid-ordinate rule : —
(a) Taking 5 ordinates only, we have the values —
X
4#2+7*- 5
y
'i
2*
3t
4i
5i
9+10-5 — 5
25 + I7-5-5
49+24-5-5
81+31-5-5
121 + 38-5-5
14-5
37-5
685
107-5
154-5
Their sum = 382-5
and the average = 76-5.
(b) Taking 10 ordinates, viz., those at x = ij, if, 2|, etc.,
the values of y are 10, 19-5, 31, 44-5, 60, 77-5, 97, 118-5, 142
and 167-5
their sum = 767-5
and the average = 76-75.
Therefore, by increasing the number of ordinates measured, a
better approximation is found.
The curve is a parabola with axis vertical, and hence Simpson's
rule should give the result accurately if 3 ordinates only are taken,
viz., at x = i, 3-5 and 6.
Thus, A = 6, M = 68-5, B = 181.
-.. 6+(4x68-5)+i8i
Hence the mean height = -
= = 76-83.
MEAN VALUES 183
If, then, the law connecting the two variables is known, the
mean value of the one over any range of the other can be found
by integrating the former with regard to the latter between the
proper limits and then dividing by the range; or to express in
symbols, if y = f(x), the mean value of y, as x ranges from a to b,
is given by —
fb ,
y dx
f
J "
b-a
Example 2. — Find the mean value of e5x between x = -2 and x =
/•? g .
•a
m.v. =
7-'2 '5 t-5
2
~ 5 '*
= ^{33-12-272}
= 12-16
i. e., if the curve y = e5x were plotted between x = -2 and x = 7 its
mean ordinate would be 12-16 units.
Example 3. — If C = 5 sin 3^, find the m.v. of C, when —
, > , . , 27C .
(a) t vanes from o to —
3
(b) t varies from o to -.
Whenever dealing with the integration of trigonometric functions
it is advisable first to determine the period of the function, since
much numerical work may often be saved in this way. The sine and
cosine curves are curves symmetrical about the axis of the I.V.
(i, e., x or t, as the case may be) ; hence there is as much area above
this axis as below it, if a full period or a multiple of periods be
considered. Therefore, regarding the area above the x axis as positive
and the area below this axis as negative, the net area over the full
period is zero, so that the mean height of the curve,* and therefore
the m.v. of the function, must be zero.
For the case of C = 5 sin 3/, the period = ^ — ^ = —
coeff <•. of / 3
and hence if the m.v. is required for t ranging fiom o to — '•
. 4^
o to — , etc.
3
the result is zero.
184 MATHEMATICS FOR ENGINEERS
Hence, whenever the analysis shows that the full period is involved,
there is no need to go through the process of integration. In this
case, however, the integration is performed for purposes of verification.
27T 27T
, _ fa 5 sin -$tdt 3 x 5 / i ,\-
(a) m.v. of C. = / * 3 = - — - ( — cos 3/
J o 271 27T \ 3 - / 1
3
- -o 3
— — - (COS 27C — COS O)
27T V
= — — ('-I)
27U V '
IT
/i.\ t r fissinitdt
(b) m.v. of C. = / - =
Jo n
3
= — - (COS 7T — COS O)
7T V
5 / \ I0
= *-* (— r — i) = — .
7T V 7T
Comparing this with the amplitude, which is the maximum ordinate
of the curve and has the value 5 sin = 5, we note that —
Mean ordinate for ^ period 10 _ 2 _
Maximum ordinate for \ period ~~ TT X 5 ~~ n ~ ''
The average height of a sine curve is always -637 x the maximum
height.
Example 4. — If an alternating electric current is given by the
relation —
C = -5 sin i207T/+-o6 sin 6007^
find the mean value of C.
The graph is of great assistance in this evaluation ; and
consequently the curves cx — -5 sin 1207^, c2 = -06 sin GOOTT/, and
C = ct + c2 = -5 sin i2O7t/+-o6 sin 6007^ are plotted in Fig. 40. The
period of c, = -5 sin I2O7T/ is or >-, and of c» = -06 sin 6007^ is
I2O7C 6O
- 01 - -, so that the period of the compound curve must be ^-.
60O7T 30O 6O
From the pievious reasoning it is seen that the mean height of the
curve ct = -5 sin i207r/ must be -637 x amplitude = -637 X -5 = -3185;
and the mean height of the curve c2 = -06 sin 6007^ considered over
the same period, viz., o to , must be the mean height of the wave A,
120
since the positive and negative areas are otherwise balanced, but this
MEAN VALUES
185
must be spread over five times its usual base ; now the mean height
of the wave A is '637 X -06, so that the mean height of the curve
c2 — -06 sin 6oo7c/ over the period o to
• -637 x -06
is — — or -0076.
120 5
Then the mean value of C = -3185 + -0076 = -3261.
The nature of the average should be clearly understood ; for it
is possible that the same quantity may have two different averages
according to the way those averages are considered. Suppose a
piston is pushed by a variable force; then the average value of
that force might be found by taking readings of it at every foot
of the stroke and dividing by the number of readings taken, in
£40 X^ £ I£O
i- -Oesin gOOTTf
FIG. 40. — Problem on Alternating Current.
which case the average would be termed a space average ; or the
force might be measured at equal intervals of time, whence the
time average of the force would result.
To take another instance : — Suppose a bullet penetrates a
target to a depth of s ft. ; the average value of the force calculated
from the formula Ps =
mv*
where P is the force, would be a
mv
space average, but P calculated from the formula Pt= — , i. e., the
6
force causing change of momentum in a definite time, would be a
time average.
To illustrate further : A body starts from rest and its speed
increases at the uniform rate of 4 ft. per sec.2. Find the time and
space averages of the velocity if the motion is considered to take
186 MATHEMATICS FOR ENGINEERS
place for 6 sees. {Use the relations s = - atz, v = at and v2 = 2as.
s = - atz = - x 4 X 62 = 72
2 2
and if the limits of t are o and 6
then the limits of s are o and 72.
To find the time average of the velocity — •
8
atdt
"6^0 ~
/•e re
I vdt c
•f 0 __. * 0 J 0
= 12 f.p.S.
To find the space average of the velocity — •
V2 = 20S = 8s
or v= V&Vs
and the mean value of v = mean value of V8
/ V»SarfS i ,-(2 sV2
Hence the space average = /T/T_rt — ~^x^^\^sl?j
72
= =
Example 5. — The electrical resistance R« of a rheostat at temperature
t° C. is given by —
RJ = 38(1 + -004^).
Find its average resistance as t varies from 10° to 40° C.
(This will be a temperature average.)
m.v. =
30
38x33
30
41-8.
MEAN VALUES 187
Example 6. — If V = V0 sin qt and C = C0 sin (qt—c), find the
average value oi the power, i. e., the average value of VC.
VC = V0 sin qt . C0 (sin qt—c)
= ° °{2 sin qt . sin (qt— c)}
V C
= — °— °{COS C — COS (20* — C)}
also the period = — .
Hence the m.v. of VC
r
[cos c— cos (zqt—c)}dt
ex/-- sin (zqt-c]\q
20 '\
cos
c-— sin (4TC-c)-o+— sin (-c) j
20 20 V 'J
275 2 0 20 20
_ VoCo? 27c |~for sin (4*— e) = sin (— e)~|
— A - COS C , • / \ • * / \
47T q Land — sin (— c)+sin (— c) = o
= ~V0C0 cos c
i. e., the mean value of the power = one-half the products of the
maximum values or amplitudes with the cosine of the lag.
This is a most important result.
If c = 90°, i. e., if the lag is -, then cos c — o and the mean value
of the watts = - V0C0 X o = o ; this being spoken of as the case of
wattless current.
Exercises 16. — On Mean Values.
1. If a gas expands so as to follow the law pv — 120, find the
average pressure between the volumes 2 and 4.
2. Find the mean value of e2'5v as y varies from o to -4.
3. The mean height of the curve y = 3*3+5#— 7 is required
between the limits x — — 2 and x — +3; find this height.
4. Find the mean value of 2-i8sin](3/— 1-6) as / ranges from
•14 to 1-6.
5. What is the mean value of 4-5 +2 sin 6o/, t ranging from
o to — ? Discuss this question from its graphic aspect.
i88
MATHEMATICS FOR ENGINEERS
6. Find the mean value of p, when pv1'37 = 550, for the range of
v from 4 to 22. .
7. The illumination I (foot candles) of a single arc placed 22 ft.
above the ground, at d ft. from the foot of the lamp is given by
I = i -4 — -oid. Find the mean illumination as d varies from J ft.
to 10 ft.
8. An alternating current is given by C = -2 sin IOOTT/+-OI sin 300^.
Find the mean value of C for the range of t, o to -02 sec.
9. Taking the figures in Question 8, find the mean value of C when
t ranges from o to -01 sec.
10. Find the mean value of 5 sin 6t X 220 sin ^t, t ranging from
7C
o to -.
3
11. The table gives the values of the side thrust on the piston of
a 160 H.P. Mercedes aero engine for different positions of the crank;
the positive values being the thrust on the right-hand wall, and the
negative values being the thrust on the left-hand wall of the piston.
Angle of crank froml
top dead centre |
Total side thrust)
(Ibs.) /
O
40
80
IOO
120
140
160
1 80
20O
240
280
300
320
330
o
+ 210
O
-170
-175
-I85
— IOO
0
+95
+240
+ IOO
o
-30
o
345
360
10
40
80
1 20
160
1 80
2OO
2 2O
240
260
280
290
320
360
+20
o
—600
-*•
—720
-580
-170
0
+ QO
+ 170
+160
+ 150
+50
0
—210
0
Plot these values (treating them all as positive) and thence
determine the mean side thrust throughout the cycle.
Root Mean Square Values. — A direct electric current maybe
measured by its three effects— chemical decomposition, magnetic
effect or heating effect. An alternating current (A.C.), however,
flowing first in one direction and then in the reverse, cannot be
measured by either the first or the second of these effects, because
the effect due to the flow in one direction would be neutralised by
that due to the opposite flow ; hence an A.C. must be measured
by its heating effect.
The heating effect of a current expressed by the heat units H
may be measured by —
H =
where C is the current,
so that it will be seen that H oc C2.
The measuring instruments are graduated to give the root of
the mean value of this heating effect, i. e., the square root of the
ROOT MEAN SQUARE VALUES
189
mean value of the squares (called the root mean square value and
written R.M.S.) ; in other words, the instrument records what are
termed virtual amperes, a virtual ampere being the current that
produces the same heat in a resistance as a steady or direct current
of i ampere in the same time.
In place of the measurement of the current by its heating
effect, the ammeter might be of the electro-dynamometer type,
and in such the instrument records the mean value of C2, and the
square root of this value is called the effective current.
Similar remarks apply also to the measurement of alternating
E.M.F.
It is therefore necessary to determine R.M.S. values of functions
likely to be encountered, and to compare the virtual with the
steady.
While the determination of R.M.S. values is of greatest import-
ance from the application to electrical problems, it is also occasion-
ally of use in problems of mechanics ; thus the calculation of what
is known as a swing radius (see p. 240) is in reality a determination
of a R.M.S. value.
In order to convey the full interpretation of the term R.M.S.
value, we shall discuss first a simple arithmetical example, then
some algebraic examples, leading up finally to the trigonometric
functions.
Example j. — The values of an alternating electric current at various
times are given in the table —
t
0
•01
•02
•03
•04
•05
•06
•07
•08
•09
•i
C
o
1-8
3'5
5
6-1
6-8
6-1
5
3'5
1-8
o
Find the mean value and also the R.M.S. value of the current,
and compare the two values.
The mean value, as before explained, is the average of the given
values and is 3-96.
We must now tabulate the values of the squares; thus —
o 3-24 12-25 25 37-21 46-24 37-21 25 12-25 3'24 and o.
The sum of the squares = 201-64
the mean of the squares or M.S. =
for we must only reckon the end values as half-values when adding
IQO
MATHEMATICS FOR ENGINEERS
up the ordinates, since the end values belong equally to the sequences
on either side.
Thus — M.S. = 20-16
and the square root of the result = R.M.S. = V2o-i6 = 4-5.
This question might have been worked entirely by graphic methods,
according to the following plan —
Plot the values of C to a base of /, giving the curve ABD in Fig. 41 ;
find the area under the curve and divide by the base, thus obtaining
the mean height; plot also the values of C2 against those of t, giving
the curve EFG. By graphic summation determine the area under
the curve EFG and draw the line MM at the mean height of the
diagram. Make MN = i unit on the scale of C2 and on PN describe
a semicircle ; produce MM to cut the semicircle in R. Then MR = 4-5
is the R.M.S. value.
Thus the mean value = 3-96 and the R.M.S. value = 4-5, and the
,. R.M.S. value 4-5
ratio - — - - = -2-i = 1-1
mean value 3-96 —
factor.
., /•
this ratio being termed the form
a _
FIG. 41. — R.M.S. Value of an Alternating Current.
Example 8. — Find the R.M.S. value of the function 2X1'5— 3* as x
ranges from 2 to 5.
The square = (2xl'5—^x)2 = 4
Explanation.
dx
and the M.S. =
3
_ i
~ 3
= 13-27-
Hence the R.M.S. value —
5-2
-959)-(i6+24-38-8)]
= Vi3'27 = 3-64-
No.
Log.
5
3-43
959
2
3-43
38-8
•6990
3'5
34950
20970
2-44650
•5353
2-9818
•3010
3'5
15050
9030
1-05350
•5353
V5883
ROOT MEAN SQUARE VALUES
191
Example 9. — Suppose that an alternating electric current at any
time follows the sine law, i. e. —
C = C0 sin qt
where C is the instantaneous value of the current at any time t, and
C0 is the maximum value of the current.
Find the R.M.S. value of the current.
As we have already seen, the determination of the R.M.S. value
implies that first the square of the function at various times must
be calculated, then the mean value of these squares found, and
finally the square root of this average extracted.
To assist in the study of this important problem, the curve
y — sin x, the simple sine curve, is shown in Fig. 42, and also the
FIG. 42. — R.M.S. Value of an Alternating Current.
curve y — sin2 x, which is obtained from the former curve by squaring
its ordinates. It will be observed that whilst for the curve y = sin x
there are both positive and negative ordinates, in the case of the
curve of squares all the ordinates 'are positive. Also the period of
the curve of squares is noted to be one-half that of the simple curve;
and therefore when calculating the mean height of this curve, it is
immaterial whether the full or the half period of the simple sine curve
is taken as the range.
In the case with which we are here particularly concerned the
C 2 2Tc
square = C2 = C02 sin2 qt = — (i — cos 2qt), and the period = — , so
that the integration may be performed either over the range o to -
or o to -.
192 MATHEMATICS FOR ENGINEERS
Taking the latter range —
•a
I q C02 sin2 qt dt
Mean of the squares = M.S. = —
q
= & fq (i -cos 2qt) dt*
7C Jo
2q
oC02 TTC I .in
= *— - --- sin 27c— o-{ — sin o
27T \_ 2 2q J
or R.M.S. value = -707 x maximum value
i. e., virtual value of current or E.M.F. = -707 x maximum value of
current or E.M.F.
Hence if a meter registers 10 amps., the maximum current is - ,
•707
i. e., 14-14 amps., or there is a variation between + 14-14 and —14-14.
* In the evaluation of the integral tq (i — cos iqf) dt, it should be
it ir
noted that it can be written q dt— I9 cos -zqtdt, and the value of
Jo Jo
the second term is zero, because it is the area under a cosine curve
taken over its full period. Hence the integral reduces to / 9 dt, and
there is no need to say anything further about the second term.
Example 10. — Find the R.M.S. value of a-\-b sin 4*.
For this function the peiiod = — — -,
4 2
Then S = (a-\-b sin 4/)2 = «2+62 sin2 4^ + 2^6 sin 4^
52
= «2+— (i — cos 8t)-\-2ab sin 4^.
7
2 fa 6a 62
Hence M.S. = - I (az-\ ---- cos 8t+zab sin 4*)
7T J 0 22
_ TT n
2 F T2 / fc2\ T2 fe2
= ~ I \aZ+2) dt—J -£
dt
cos tt + 2a sn
VOLUMES 193
2 2 2
R.M.S.=
Exercises 17.— On Root Mean Square Values.
Find the R.M.S. values of the functions in Nos. i to 7.
1. #3+2 (x ranging from i to 3).
2. e~"x (x ranging from — -i to +-65).
3. 3-4 sin 5*i/. 4. -165 cos (-07— 2/).
5. 1-4 tan 2t (t ranging from o to -43).
6. i -14 +-5 cos -8t.
7. -72 sin (3— 4/) ; compare with the mean value.
o -c- A .LI. t * /R.M.S. value\ , ,,
8. Find the form factor — -, — ) of the wave —
\ mean value /
e = Ex sin pt+~Ea sin 3pt.
9. Compare the " effective " values of two currents, one whose
wave form is sinusoidal, having a maximum of 100 amperes, and another
of triangular shape with a maximum of 150 amperes.
10. An A.C. has the following values at equal intervals of time :
3. 4. 4'5, 5'5. 8, io- 6, o, —3, —4, —4-5, —5-5, —8, — 10, —6, p.
Find the R.M.S. value of this current.
11. A number of equal masses are attached to the ends of rods
rotating about one axis. If the lengths of these rods are 10, 9, 5,
8, 4, 13 and 15 ins. respectively, find the effective radius (called the
swing radius) of the system. (This is the R.M.S. value of the respective
radii.)
12. Find the R.M.S. value of the function sin2 0 cos3 6 over the
period o to -. (Refer to p. 178.)
13. The value of the primary current through a transformer at
equal time intervals was —
•20 -05 -07 -ii -14 -19 -2i -04 -08 '12 -15 -18 -2i -08 -04
Find the R.M.S. value of this current.
Volumes. — If a curve be drawn, the ordinates being the values
of the cross sections of a solid at the various points along its
length, then the area under the curve will represent, to some
scale, the volume of the solid. For, considering a small element
of the length, 81, if the mean area over that element is A, the
o
194 MATHEMATICS FOR ENGINEERS
volume SV of the small portion of the solid is A.U, and the
total volume of the solid is the sum of all such small volumes,
i. e., 2AS£. If the length 81 is diminished until infinitely small,
2AS/ becomes fAdl, and hence —
Volume of solid = / Adi.
J o
Comparing this with the formula giving the area of a closed
figure, viz.,fydx, it is seen to be of the same form, and it is made
identical if A is written in place of y (i. e., areas must be plotted
vertically) and / in place of x (i. e., lengths must be plotted
horizontally) .
When values of A and / are given the curve should be drawn
and integrated by any of the given methods ; and if it is preferred
to find the actual area of the figure in sq. ins. first, the number
of units of volume represented will be deduced from a consideration
of the scales used in the plotting. Thus if i" (horizontally) repre-
sents x ft. of length, and i" (vertically) represents y sq. ft. of area,
then i sq. in. of area represents xy cu. ft. of volume. To take
a numerical example : If i" = 15" of length, and i" = 5 sq. ft. of
area, then i sq. in. of area = — x 5 = 6J cu. ft. of volume.
12
If the area is found by the sum curve this conversion is
unnecessary, as the scales are settled in the course of the drawing.
The Example on p. 122 is an illustration of the determination of
volumes by graphic integration; in that case the actual areas of
sections are not plotted directly, but values of d2, the multiplication
by the constant factor being left until the end. Had the solid not
been of circular section the actual areas would have been plotted
as ordinates and the work carried on as there detailed.
We thus see that the determination of the volume of any
irregular solid can be effected, if the cross sections at various
distances from the ends can be found, by a process of graphic
integration.
If, however, the law governing the variation in section is known,
it may be more direct to perform the integration by algebraic
methods.
To take a very simple illustration : —
Example n. — The cross section of a certain body is always equal
to (5#3 + 8) sq. ft., where x ft. is the distance of the section from one
end. If the length of the body is 5 ft., find its volume.
VOLUMES 195
The body might have an elevation like Fig. 43, and its cross section
might be of any shape ; the only condition to be satisfied being that
the area of a cross section such as that at BB must = 5*3+8. Thus
the area at AA must = (5 X o) + 8 = 8 and area at CC = (5 X 53) + 8 = 633.
f5 C5
Then the volume = J QAdx = ] Q(5x3+8)dx
+40 = 821-25 cu. ft.
~"~^^
4
Volumes of Solids of Revolution. — A solid of revolution is
generated by the revolution of some closed figure round an axis
which does not cut the figure. Thus, dealing with familiar solids,
the right circular cone, the cylinder and the sphere are solids of
revolution, being generated respectively by the revolution of a
right-angled triangle about one of the sides
including the right angle, a rectangle about
one of its sides, and a semicircle about its
diameter ; and of the less well-known solids
of revolution the most important to the
engineer is the hyperboloid of revolution
which is generated by the revolution of a FIG
hyperbola about one of its axes and occurs
in the design of skew wheels. The axis about which the revolution
is made in all these examples lies along a boundary of the re-
volving figure; whereas an anchor ring is generated by the
revolution of a circle about an axis parallel to a diameter but
some distance from it.
The revolving figure may have any shape whatever, the only
conditions being, for the following rule to hold, that the axis about
which the revolution is made does not cut the figure and that the
cross section perpendicular to this axis is always circular.
Imagine the revolving cross section to be of the character
shown in the sketch (ABCD in Fig. 44) ; and let the revolution
be about the axis of x. It is required to find the volume of the
solid of revolution generated.
Working entirely from first principles, i, e., reverting to our
idea of dealing with a small element and then summing : if the
strip MN of height y and thickness 8x revolves about OX it will
generate a cylinder.
The radius of this cylinder will be y and its height or length Sx ;
and hence its volume = iry2Sx.
Accordingly the volume swept out by the revolution of ABCD
ig6
MATHEMATICS FOR ENGINEERS
will be ^nyzSx (the proper limits being assigned to x) approximately,
or fny2dx accurately.
Again, it will be seen that such a volume can be measured by
the area of a figure, for, writing Y in place of Try2, the volume
= fYdx, which is the standard form for an area. Hence, if
values of y are given, corresponding values of Try2 must be cal-
culated and plotted as ordinates, and the area of the resulting
figure found.
It should be noted that fny2dx might be written as itfy*dxt
thus saving labour by reserving the multiplication by TC until the
M
B X
X!
FIG. 44. — Volume of Solid of Revolution.
area has been found, i. e., the values of y2 and not those of Try2
are plotted as ordinates.
The following example will illustrate : —
Example 12. — The curve given by the figures in the table revolves
about the axis of x; find the volume of the solid generated, the
bounding planes being those through x = 2 and x = j, perpendicular
to the axis of revolution.
X
2
3
4
5
6
7
y
44
42
44
46
45
38
VOLUMES 197
n
Values of y2 must first be calculated, since the volume = / jtyzdx
=
Hence the table for plotting reads —
X
2
3
4
5
6
7 •
y*
1936
1764
1936
2116
2025
1444
and the values of y2 are plotted vertically, the curve ABC (Fig. 45)
resulting.
This curve is next integrated from the axis of x as base, the curve
DEF resulting; the polar distance being taken as 3, so that the new
O 6 /
FIG. 45. — Volume of Solid of Revolution.
vertical scale = 3 X old vertical scale. It must be remembered, how-
ever, that the base from which the summation has been made in the
figure is not the true base, since the first value of the ordinate is 1400
and not o; thus a rectangle 1400X5 has been omitted. Hence we
must start to number our scale at 7000 ; and according to this
f7
numbering the last ordinate reads 9580, hence / zyzdx = 9580, or —
/•?
Volume = TC I yzdx
= TT x 9580
= 30,100 cu. ins.
In cases in which y and x are connected by a law the integra-
198
MATHEMATICS FOR ENGINEERS
tion may be performed in accordance with the rules for the
integration of functions.
Example 13. — Find the volume of a
paraboloid of revolution and compare it
with the volume of the circumscribing
cylinder.
o
Y
B
FIG. 46.
A paraboloid of revolution is generated
by the revolution of a parabola about
its axis. Suppose the parabola is placed
as shown in Fig. 46; the revolution is
therefore about OX.
The equation to the curve OB is
y2 = $ax (see Part I, p. 106), i. e., if
OA = h, (AB)2 = 4ah.
Hence the volume of the solid swept
out by the revolution of OBA —
[h [h
= Ttyzdx = TU I qax dx
J J 0
_ 4a7L><Aa _ 7,2
Now the volume of the circumscribing cylinder —
= TC X 4ah X h
and hence the volume of the paraboloid —
= - X vol. of circumscribing cylinder.
Example 14.— The curve y2 = 64 — zx2 revolves about the axis of y.
Find the volume of the solid generated, the limits to be applied to x
being o and 5.
This differs from the cases previously treated in that the revolution
is to be about the y axis and not about the x axis.
Hence the volume = fnx2dy and not Jwy*dx, the y replacing x
and vice versa. Also another point must be noted : the limits given
are those for x, whereas the limits in the integral fiix2dy must apply
to the I.V., which is now y. Therefore a preliminary calculation
must be made to determine the corresponding limits of y —
y2 = 64-
x = o, y
x = 5,
2 = 64, y = ±8
= 14- y = ±374-
VOLUMES
199
The double signs occurring here may possibly confuse, but actually
the equation given is that of an ellipse, symmetrical about the axes
of x and y, and the volume required is the volume generated by the
revolution of the two shaded portions (Fig. 47), which will be twice
that generated by one of these; hence, taking the upper shaded
portion, we use the positive limits, viz., 3-74 and 8.
f8 f8 / v2\
Then the volume = I r:xzdy — ni ( 32— - }dy
J 3-74 ' J 3-ll\J 2 / '
f y3\8
= m say— if )
6 /3-74
= 7c(256— 85-3-H9-5 + 8-7)
= 59 '9^.
The solid due to the revolution of the lower portion will be also 59-97^,
and hence the total volume generated = 119-871; = 376 cu. units.
FIG. 47.
If the limits for y were —8 and +8, the volume of the whole solid
would be required ; then —
[8 f8/ y2\ r y3-|3
Volume = / -KXzdy — 2:rJ \^)^ — — ]dy = 2TC[_32)'— z~J
- 27^(256-85-3)
= 341-471 = 1070 cu. units.
The solid generated by the revolution of an ellipse about its
major axis is known as a prolate spheroid; while if the revolution
is about the minor axis the solid is an oblate spheroid.
The volumes of these may be necessary, so that they are given
in a general form.
200
MATHEMATICS FOR ENGINEERS
The general equation of an ellipse is —2-\-jp = i (cf.
P- 344)-
Let a >£>, i. e., the major axis is horizontal.
For a prolate spheroid the revolution is about the major axis,
/a fa J2
nyzdx = 2rr/ ~-2(az—xz)dx
o J o a
27C&V „
FIG. 48.
In like manner, the volume of an oblate spheroid = - na?b ;
and it should be noted that if b = a, the spheroid becomes a
sphere and its volume = ~KO.S.
3
Example 15. — Find the volume of a zone of a sphere of radius r,
the bounding planes being those through x = a and x = b.
The equation of the circle is xz+yz = r2 (Fig. 48)
whence y2 = rz—xz.
:. Volume of a zone = fbny2dx = TT fb(rz—x2)dx
J a J a
= 7r[»-2(6-a)--(63-a3)]
LENGTH OF ARC
20 1
This can be put in the form given on p. 120, Part I, if for (b—a)
we write h, and for BE and DF their respective values r: and r2.
Thus az = rz-r]z, bz = rz—rzz
hz = (b-a)z = bz+az—2ab = rz—r1z+rz-r2z-2ab
, — hz+2rz—r1z—r2z
and hence ab = — -
2
So that the volume of the zone —
Length of [Arc. — Consider a small portion of a curve, PQ in
Fig. 49, P and Q being points near to one another.
FIG. 49.
The small length of arc PQ is denoted by 8s, so that a complete
arc would be denoted by s.
Let PM = 8*, and QM = Sy.
Then the arc PQ = the chord PQ very nearly, so that we may
say —
(Ss)2 = (chord PQ)2 = (S#)2+(8y)2.
/Ss\2
•• y =i
*• £•> v~ =
Hence when 8x becomes infinitely small, , - becomes ,-, etc.
J 8x dx
and
ds I
J- = A/
dx \
dy
ds
or - =
dx
202 MATHEMATICS FOR ENGINEERS
Integrating — •
dz
dx °r
The length of arc can thus be found if the value of the integral
on the R.H.S. can be evaluated.
In only a few cases is the evaluation of the integral simple;
and for most curves an approximation is taken, e. g., to find the
perimeter of an ellipse by this method one would become involved
in a most difficult integral known as an elliptic integral, this being
treated later in the chapter; and hence the approximate rules are
nearly always used in practice.
To deal with a case of a very simple character :—
Example 16. — If y — ax-j-b, find the length of arc between x — m
and x = n.
In this case it is really a matter of determining the length of the
line AB (Fig. 50) ; the slope of the line being a.
Then
if y = ax+b, = a and i + = i+«2.
/m . __
Vi+a2dx =
n
= Vi + a2 (m — ri).
On reference to the figure it will be seen that this is a true result,
since AB = \/(AC)2+(CB)2 = V(m-w)2+a2(w-w)2
= Vi+a2 (m — n).
Example 17. — Use this method to determine the approximate
length of a cable hanging in a parabola, when the droop is D and
the span is 2L.
For convenience, put the figure in the form of Fig. 51.
Then L2 = 4«D
L2
whence a = -~, so that a must be very large.
LENGTH OF ARC 203
The equation of the curve is in reality —
y being written in place of L, and x in place of D.
men
dx
d
— = -5 — A ~r~
x ay dx
dy
JC
so that
or
-/-• 2V
dx
dy
dx
= 4<z
AO, 2a
~ zy ~ y u
^L^
whence
dx
- Z.
T
dy
20,'
i
Thus
FIG. 51.
V
= 1 + 5-2 approximately,
O#
since all the subsequent terms contain a4 and higher powers of a in
the denominator, so that all these terms must be very small.
Hence s = 2 1 + . dy
-
.
3 Span
Example 18. — Find the length of the circumference of a circle of
radius r.
The equation of the circle is y2+#2 = y2.
Thus y2 = rz-x2
and ay 4 -^- = — a# {differentiating with regard to x}
ctx
or ^ = -- - ^
204 MATHEMATICS FOR ENGINEERS
Hence i + (;,— ) = i+~i s= -
Length of circumference
— 4 X length of J circumference
fa r
= 4 J o "V^F2 **
To evaluate this integral, let x = r sin u (cf. Example 7, p. 150).
Then r2— x2 = r2—r2 sin2 u = r2(i — sin2 «)
and also
= r* cos* u
dx
— = r cos u
du
dx = r cos u . du.
y c
FIG. 52.
To find the limits to be assigned to u —
sin u = - ; and when x = o, u = o
r
and when x = i, u = 90° or
Thus the circumference
ra dx
= *r J oVi*=x
dx
r cos u du
o r cos
7T
2
0
Example 19. — Find an expression for the length of the perimeter
of the ellipse whose major and minor axes are 2a and 26 respectively.
LENGTH OF ARC 205
Let BPA (Fig. 52) be the ellipse, CQA the quadrant of a circle
on the major axis as diameter, and BTD the quadrant of a circle on
the minor axis as diameter. Selecting P as any point on the ellipse,
draw the lines QPN, PT, TS, QR and OTQ as shown.
The point P has the co-ordinates x and y, viz., ON and PN, which
are respectively equal to QR and TS.
Now QR = OQ sin <j> or x — a sin <f>
and TS = OT cos <p or y = b cos <p.
Thus — = a cos* and ~- = —
or 5 = Va2cos2<f>+&2sin2<f> d<f>.
x* v2
Now the equation to the ellipse is ~^+hz = i, and the eccentricity,
d 0
which we shall denote by K, is given by —
T, distance between foci A/a2— b2 OF .
K — - — . ---- . — — — - — = ~--i-, F being a focus.
major axis a OA
Hence K2 = °^- and K2a2-a2 = -62
or Z>2 = a2(i-K2)
so that a*cosaf +Msin8f = a^os^+a^in2^— a2K2sin2f
= a2(i-K2sin2$>).
Thus our integral reduces to the form —
and for the quarter of the ellipse the perimeter = I a Vl— K^in2^ d$,
J o
since the limits for $ are obviously o and .
it
p _
Also the » full perimeter of the ellipse = 4 I a vi — K^in2^ d<t>.
Jo
This integral, called an " elliptic integral of the second kind," is
extremely difficult to evaluate; but in view of the importance of the
perimeter of the ellipse it is well that we should carry the work a
little further.
Knowing the values of a and K for any particular ellipse, recourse
may be made to tables of values of elliptic integrals, but if these are
not available, a graphic method presents itself which is not at all
difficult to use. According to this plan, various values of $ are
jT . _ -
chosen between o and , and the calculated values of vi — K2sin2<?>
are plotted as ordinates to a base of <f>. Then the area under the
resulting curve when multiplied by ^a gives the perimeter of the
ellipse.
206
MATHEMATICS FOR ENGINEERS
o.
ees
ur
1-6
1-5
1-4.
1-3.
1-2
1-0
0
FIG. 53.: — Perimeter of Ellipse.
—
— ^
—
—
^
Xs
=
\
\
—
\
\
—
\
—
P
^ri meter or Ellipse
2 x major axis x orainct
\
1
c
t* '
1-294
\
=
\
i
\
\
—
\
=
1
i
\
—
i
A
II
MM
MM
II
III
II
1
II
0 •/ -2 3 -4- -S -6 -7 -8 -9 /
e.cce.n/~ric.tfuK
FIG. 54.
LENGTH OF ARC
207
Example 20. — A barrier before a ticket office in a works was
constructed out of sheet metal, which was bent to the form of an
ellipse of major axis 21 ins. and minor axis 5 ins. Find the area of
sheet metal required if the height of the barrier is 5 ft.
T j-u- j tr VlO'52 — C
In this case a = 10-5 and K = —
10-5
so that K2 = -944.
The table for the plotting reads —
= '9714
4>
sin <f>
sin1 <p
l-K»sin*£
Vl-K*sin»^
o
O
0
I
I
•157
•1564
•0245
i --0232 = -977
•99
•314
•309
•0951
— 0898 = -91
•955
•471
'454
•206
— 195 = -805
•897
•628
•5878
•345
-•326 = -674
•821
•785
•7071
•5
— 472 = -528
•726
•942
•809
•652
— -616 = -384
•62
1-099
•891
•793
— 749 = -251
•501
1-256
•9511
•9
-•85 = -15
•388
i-4J3
•9877
-976
— 922 = -078
•279
1-570
i
I
— •944 = -056
•235
and the values in the extreme columns are plotted in Fig. 53.
The area under this curve = 1-0663 SCL- umt
and thus the perimeter = 40 x i — 4 X 10-5 x 1-0663 = 44-78 ins.
Hence the area required =
12
ft
^
It is well to compare this value of the perimeter with those
obtained by the approximate rules given in Part I, p. 105 —
(a) Perimeter — TU(«+&) = 71(10-5+2-5) = 40-7 ins.
(b) Perimeter = 4-443 V«2+62 = 4-443x10-8 = 47-8 ins.
(c) Perimeter = 7u{i-5(a+&) — Vab} = TTX 14-38 = 45-1 ins.
and the perimeter, correct to two places of decimals, is given in
the tables * as 44-79 ins.
* The tables of complete elliptic integrals give the values of
0 Vi — K^sin2^ d<p for various values of 0, 0 being the angle whose
sine is K, the eccentricity of the ellipse. Thus to use the tables for
this particular case we put sin 0 — K = -9714, whence 0 = 76° 16';
we then read the values of the integral for 75°, 76° and 77°, and by
plotting these values and interpolation we find that for the required
208 MATHEMATICS FOR ENGINEERS
Thus the errors in the results found by the different rules are —
(a) 9-13 % too small (b) 6-69 % too large (c) -67 % too large
showing that the rule of Boussinesq gives an extremely good result
in this case of a very flat ellipse, whilst the other approximate
methods are practically worthless.
Area of Surface of a Solid of Revolution. — When a solid
of revolution is generated, the boundary of the revolving figure
sweeps out the surface of that solid. The volume of the solid
depends upon the area of the revolving figure, whilst the surface
depends upon the perimeter of the revolving figure.
To find the surface generated by the revolution of the curve CD
about OX (Fig. 55) we must find the sum of the surfaces swept out
by small portions of the curve, such as PQ. Let PQ = a small
element of arc = 8s. Then the outside surface of the solid generated
by the revolution of the strip PQMN about OX will be equal to
the circumference of the base X slant height, i.e., 2ny8s. Hence
the total surface will be the sum of all similar elements, i. e.,
\r\x = b
> 2Tcy8s, approximately, or if 8x becomes smaller and smaller —
^L_— J iC — tt
fx = b
Surface = I 2nyds.
J x = a *
For ds we may substitute its value, viz. —
/b
2ny
(J
2nyJ !--} .dx.
angle, viz., 1-0664. Multiplication by 40, i.e., 42, gives the result
44-79. For the convenience of readers interested in this question,
and who desire a result more exact than that given by the approximate
»r
fi _
rules, a curve is here given (Fig. 54) with values of I v i — K^in2^ d$
J o
plotted against values of K; and for the full perimeter of the ellipse
the ordinates of this curve must be multiplied by twice the length
of the major axis.
E. g., if the major axis = 16 and the minor axis = 10
K = - g-^ = -7807.
Erecting an ordinate at K = -7807 to meet the curve, we read
the value 1-294; multiplying this by 32, we arrive at the figure 41-41,
which is thus the required perimeter.
AREA OF SURFACE
209
Example 21. — Find the area of the surface of a lune of a sphere
of radius a, the thickness or height of the lune being b.
The surface will be that generated by the revolution of the arc CD
of the circle about its diameter OX (Fig. 56).
From the figure y2 = a2—x2
whence 2V.-r — — 2X
dx
dy x
or —- = — .
dx y
„_ fds\2 ,(dy\2 . x2 y2+x2 a2
Thus (j- ) =i+l/) = i-f - - = Z—Zy- =
\dxJ \dx' y2 y2 a2—x2
fd a
Hence the surface = I 27uVa2— x2— — -
J c Va2-
p
- 2Tiaj
dx
= 2r:a(d—c) =
but 2TOZ& is the area of a portion of the lateral surface of the cylinder
circumscribing the sphere.
Thus the surface of a lune of a sphere = the lateral surface of the
portion of the cylinder circumscribing the sphere (the heights being
the same).
Exercises 18. — On Volumes, Areas of Surfaces and Length of Arc.
1. The cross sections at various points along a cutting are as follows —
Distance from one end (ft.)
o
40
82
103
134
1 66
192
200
Area of cross section (sq. ft.)
o
210
296
205
244
154
50
o
Find the volume of earth removed in making the cutting.
2IO
MATHEMATICS FOR ENGINEERS
2. Find the weight of the stone pillar shown in Fig. 57. The
flanges are cylindrical, whilst the radius of the body at any section
2
is determined by the rule, radius = — ;=, where x is the distance of
vx
the section from the fixed point O. (Weight of stone = 140 Ibs. per
cu. ft.)
3. The curve y = 2x2—$x revolves about the axis of x. Find
the volume of the solid thus generated, the bounding planes being
those for which x — — 2 and x = +4.
4. Find, by integration, the surface of a hemisphere of radius r.
5. The curve y = aebx passes through the points x — i, y = 3-5,
and x = 10, y = 12-6; find a and b. This curve rotates about the
axis of x, describing a surface of revolution. Find the volume between
the cross sections at x — i and x — 10.
6. Find the weight of a cylinder of length / and diameter D, the
density of the material varying as the distance from the base. (Let
the density of a layer distant x from base = K#.)
•^r-o1
FIG. 57. — Weight of Stone Pillar.
7. The rectangular hyperbola having the equation x2— yz = 25
revolves about the axis of x. Find the volume of a segment of
height 5 measured from the vertex.
8. The line 4y— 5* — 12 revolves about the axis of x. Find the
surface of the frustum of the cone thus generated, the limits of x
being i and 5.
9. The radius of a spindle weight at various points along its length
is given in the table —
Distance from one end (ins.)
o
•375
•5
i-o
i'3
1-6
1-85
1-61
1-61
•78
•42
• 4
•«
• c
Find its weight at -283 Ib. per cu. in., the end portions being
cylindrical.
10. Determine by the method indicated in Example 19, p. 204,
the perimeter of an ellipse whose major axis is 30 ins. and whose
minor axis is 18 ins. Compare your result with those obtained by
the use of the approximate rules (a), (b) and (c) on p. 207.
11. The curve taken by a freely hanging cable weighing 3 Ibs.
per foot and strained by a horizontal pull of 300 Ibs. weight conforms
to the equation —
, x
y = c cosh
300
where c — •= — .
3
Find the total length of the cable if the span is 60 ft., i. e.. x
ranges from —30 to -{-30.
B
CENTROIDS 211
Centre of Gravity and Centroid. — The Centre of Gravity
(C. of G.) of a body is that point at which the resultant of all the
forces acting on the body may be supposed to act, »'. e., it is the
balancing point. The term Centroid has been applied in place of
C. of G. when dealing with areas ; and as our work here is more con-
cerned with areas it will be convenient to adopt the term centroid.
From the definition it will be seen that the whole weight of
a body may be supposed to act at its C. of G. ; and in problems in
Mechanics this property is most useful. Thus, movements of a
complex system of weights may be reduced to the movement of
the C. of G. of these. Or to take another instance : in structural
work, in connection with fixed beams unsymmetrically loaded, it
is necessary to find the position of the centroid of the bending-
moment diagram. It is thus
extremely important that rules
should be found for fixation of
the position of the centroid in „ '
all cases ; and the methods Jt »J JL jt »&
t i , ,..,,. Tnt ff/2 "?3 III A. iffS
adopted may be divided into
two classes : (a) algebraic (in- "IG" 5 8. -Centre of Gravity or Centroid.
eluding purely algebraic, and partly algebraic and partly graphic),
(b) graphic.
The rules will best be approached by way of a simple example
on moments. In place of areas or solids, afterwards to be dealt with,
let us consider the case of a uniform bar loaded as shown in Fig. 58.
For equilibrium the two conditions to be satisfied are —
(1) The upward forces balance the downward forces.
(2) The right-hand moments about any point balance the left-
hand moments about the same point; or, in other words, the
algebraic sum of the moments about any point is zero.
If C is the balancing point or fulcrum, the upward reaction of
the fulcrum = M = fni-^-m2-}-m3-{-mt-\-m5 from condition (i).
Taking moments about A, let x (x bar) be the distance AC.
Then, by condition (2) —
or
The product of a force into its distance from a fixed point or
axis is called its first moment about that point or axis ; whilst the
212
MATHEMATICS FOR ENGINEERS
product of a force into the square of its distance from a fixed
point is called its second moment about that point.
Hence our statement concerning the distance AC can be
written — •
1st moments
x =
masses
To extend this rule to meet the case of a number of scattered
masses arranged as in Fig. 59, the co-ordinates of the centroid
must be found, viz., x and y.
Thus
FIG. 59. — Centroid.
2mx 2 1st moments about OY
and
2m 2 masses
_ ^my _ ^ 1s* moments about OX
^ "" 2m 2 masses
If the masses are not all in one plane, their C. of G. must be
found by marking their positions in a plan and elevation drawing
and determining the C. of G. of the elevations and also that
of their plan. Thus the C. of G. is located by its plan and
elevation.
It will be observed that some form of summation is necessary
for the determination of the positions of centroids or centres of
gravity; and this summation may be called by a different name,
viz., integration, all the rules of which may be utilised; the
integration in some cases being graphic, in some cases algebraic,
and in others a combination of the two.
CENTROIDS
213
Rules for the Determination of the Centroid of an
Area. — Let it be required to find the centroid of the area ABCD
in Fig. 60.
The area may be considered to be composed of an infinite
number of small elements or masses, each being the mass of some
thin strip like PQMN ; the co-ordinates of the centre of gravity of
which may be determined in the manner already explained.
._.b
N
M
B
FIG. 60. — Centroid of an Area.
To find x, i. e., the distance of the centroid from OY —
Mass of strip PQMN = area x density (considering the strip as of
unit thickness)
= y8xxp
ist moment of strip about OY = mass X distance = pySx x x
= pxySx.
ist moments about OY
Hence
x =
masses
= _/ %. the limits being a and b
_5 pyox
and if the strips are made extremely narrow—
/& fb
pxydx I xydx
a J a
X =
/pydx I ydx
a ' J a
p cancelling from both numerator and denominator.
214 MATHEMATICS FOR ENGINEERS
Thus a vertical is found on which the centroid of the area
must lie; and this line is known as the centroid vertical.
To fix the actual position of the centroid some other line must
be drawn, say a horizontal line, the intersection of which with
the centroid vertical is the centroid.
Thus the height of the centroid above OX must be found;
this being denoted by y.
To find y. — The whole mass of the strip PQMN may be supposed
to act at R, its mid-point, because the strip is of uniform density ;
and hence the moment of the strip PQMN about OX
y
= mass X distance = py 8xx-
2
2b ist moments about OX
Hence y = - ,
2, masses
b 2 fb
py dx I y dx
a. J a
As in previous cases, the integration may be algebraic, this
being so when y is stated in terms of x, or graphic, when a curve
or values of y and x are given.
Suppose the latter is the case, and we desire to find x —
T~ fxydx
Then x =JTZT-
Jydx
and the values of the numerator and denominator must be found
separately. Each of these gives the area of a figure, for if Y is
written in place of xy, the numerator becomes fYdx, which is
the standard expression for the area under the curve in which Y
is plotted against x; and the denominator is already in the
required form.
Thus a new set of values must be calculated, viz., those of Y,
these being obtained by multiplication together of corresponding
values of x and y; and these values of Y are plotted to a base
of x. Then the area under the curve so obtained is the value of
the numerator, and the denominator is the area under the curve
with y plotted against x; and, finally, division of the one by the
other fixes the value of x.
CENTROIDS
215
Example 22. — Find the centroid of the area bounded by the curve
given by the table, the axis of x and the ordinates through x — 10
and x = 60.
X
IO
25
40
45
5«
60
y
4
5'3
6-2
6-4
6-6
6-8
We thus wish to find the centroid of the area ABCD (Fig.
To find x : —
xy
30Q
toa
C e ntroid Horn
ntal.
'
JS
JG D
ZO 3O 4O
FIG. 61. — Centroid of an Area.
The table for the plotting of Y against x reads —
X
IO
25
40
45
50
.
60
Y or xy
40
132-4
248
288
330
408
From this we get the curve AEF.
The area of the figure ABCD—
/GO
ydx = 289
10
and the area of the figure AEFD —
/GO
xydx = 10650
10
The method
of integration
is not shown,
to avoid con-
fusion of
curves.
/GO
xydx
10
r60
/ ydx
J 10
10650
289
= 36-9.
2l6
MATHEMATICS FOR ENGINEERS
Thus the centroid vertical, or the line PG is fixed.
We need now to find the centroid horizontal, i. e., y must be
determined.
i r60
= 2 J10Y^ /where Y in this case\
areaofABCD \ stands for y2 J
Now
y =
f° rf
J 10
so that the following table must be compiled —
0
X
10
25
40
45
5«
60
Y or y2 .
16
28
38-4
4i
43-5
46-1
FIG. 62. — C. of G. of Thin Plate.
Plotting from this table, the curve RQ results, and the area of
the figure ARQD is 1689.
_ I area of ARQD &xi68Q
y ": areaofABCD : 289 ~ ^22i
The intersection of the centroid vertical and the centroid horizontal
at G fixes the centroid of ABCD.
A modification of this method is necessary when the actual
area is given in place of the tabulated list of values, the procedure
being outlined in the following example.
Example 23. — It is required to find the C. of G. of a thin plate
having the shape shown in Fig. 62. Show how this may be done.
Draw two convenient axes at right angles and divide up the area
into thin strips by lines drawn parallel to OY. Draw in, also, the
mid-ordinates of these strips. The area of any strip can be assumed
CENTROIDS
217
to be "mean height X thickness "; and therefore measure ordinates
such as MN and multiply by the thickness or width of the strip.
Repeat for each strip, and the sum of all these will be the area of the
figure.
To find x. — OAX = the distance of the centre of ist strip from OY
so that the area of strip X OAV = ist moment of strip about OY.
Hence, multiply the area of each strip by the distance of its mid-
ordinate from OY and add the results ; then the sum will be the
ist moment of the area about OY.
T,, - Sum of ist moments 2nd total
Then x = r — = — . . . ..
Area ist total
To find y. — Fix R, the mid-point of MN, and do the same for
all the strips. The area of the strip has already been found ; multiply
this by AjR and repeat for all strips. The sum of all such will be
the ist moment about OX; dividing this by the area of the figure,
the distance, y, of the centroid from OX is found.
[Note that R is the mid-point of MN and not of NAX, because
OX is a purely arbitrary axis.]
For this example the calculation would be set out thus —
Strip
Length of
mid-ordinate
(like MN)
Width
of
Strip
Area
of
Strip
Distance of
centre from
OY
(like OAj)
Distance of
centre from
OX
(like RAi)
ist
moment
about
OX
ist
moment
about
OY
I
i-55
•5
•775
•25
2-0
1-55
•19
2
2-79
•5
1-395
•75
2-O
2-79
1-05
3
3'44
•5
1-720
1-25
2-18
3-75
2-05
4
3-85
•5
1-925
i-75
2-27
4*37
3-36
5
4-01
•5
2-005
2-25
2-32
4-64
4-50
6
3-92
•5
1-960
2-75
2-3
4-5i
5-40
7
3-60
•5
i -800
3-25
2-18
3-92
5-85
8
3-26
•5
1-630
3-75
2-04
3-32
6-II
9
2-66
•5
i-33o
4-25
2'*O2
2-68
5-65
10
1-47
•5
•735
4-75
1-95
i-43
3*49
Totals
I5-275
32-96
37-65
and
15-
= 6
32-96
y = - -— — 2-16.
15-28
Thus the position of G is fixed by the intersection of a horizontal
at a height of 2-16 with a vertical 2-46 units distance from OY.
If the centroid of an arc was required, the lengths of small
elements of arc would be dealt with in place of the small areas,
but otherwise the procedure would be the same.
2l8
MATHEMATICS FOR ENGINEERS
"Double Sum Curve " Method of Finding the Centroid
Vertical. — This method is convenient when only the centroid
vertical is required ; for although entirely graphic, it is rather too
long to be used for fixing the centroid definitely.
Method of Procedure. — To find the centroid vertical for the area
APQH (Fig. 63).
Sum curve the curve PQ in the ordinary way, thus obtaining
the curve AegE; for this construction the pole is at O, and the
polar distance is p.
Produce PA to Ol7 making the polar distance pt = HE = last
ordinate of the sum curve of the original curve (viz., PQ).
Sum curve the curve AegE from AP as base and with Ot as
FIG. 63. — CentroidiVertical~of~an Area.
pole ; then the last ordinate of this curve, viz., CM, is of length x,
so that the vertical through C is the centroid vertical.
Proof. — Consider the strip abed, a portion of the original area.
Then Or and eg are parallel (by construction) —
P ef ab
-*- = - / = -f
Ar fg fg
and thus
or
i.e., hnxab =
2hn xab = 2p xfg = pSfg = p . HE.
Again, the ist moment of the strip about AP = area X distance
= hn x ab X Ah
m = hnxabxml
= pXfgXml
CENTROIDS
219
and hence ist moment of area APQH about AP
= plfgxtnl
but ist moment of area APQH about AP —
= area x distance of centroid from A
and
x = MC.
fee/.
FIG. 64. — Problem on Loaded Beam.
Example 24. — A beam, 16 ft. long, simply supported at its ends
is loaded with a continuously varying load, the loading being as
expressed in the table.
Distance from left-hand)
support (feet) /
o
2
4
6
8
10
12
14
16
Load in tons per foot run
•12
•17
•21
•25
•28
•29
•3*
•34
•38
Find the centroid vertical of the load curve, and hence determine
the reactions of the supports and the point at which the maximum
bending moment occurs.
We first plot the load curve from the figures given in the table
(Fig. 64) ; and next we sum curve this curve, taking a polar distance
of 10 horizontal units ; the last ordinate of this sum curve reads 4-27,
so that the total load is 4-27 tons. We now set off AD equal in
length to BC, and with this as polar distance we sum curve the curve
AEC fiom the vertical axis as base. This sum curve finishes at the
point G on the horizontal through C, and a vertical through G is the
centroid vertical, distant 9-2 ft. from the end A.
For purposes of calculation, the whole load may be supposed to
220 MATHEMATICS FOR ENGINEERS
act in this line; the total load is 4-27 tons, and taking moments
round A —
4-27X9-2 = RBx 16
whence RB = 2-46 tons
and RA = 4-27—2-46 = 1-81 tons.
We now set up AH, a distance to represent RA, to the new vertical
scale, and then a horizontal through H is the true base line of shear.
At the point P the shear is zero; but the shear is measured by
the rate of change of bending moment, so that zero shear corresponds
to maximum bending moment; and hence, grouping our results — •
Reaction at left-hand support = i 81 tons
Reaction at right-hand support = 2-46 tons
and the maximum bending moment occurs at a distance of 8-4 ft.
from the left-hand end.
Centroids of Sections by Calculation (for a graphic method
especially applicable to these, see p. 251). — Special cases arise in
N, r
L \(
r IfcHJ |S|I
L
N-
-NT
N2
••*
I
T
T
i.
•*I^N-
FlG. 65.
the form of sections of beams, joists, rails, etc., for which a
modification of the previous methods is sufficient.
If the section is composed of a combination of simple figures,
such as rectangles or circles, as in the great majority of cases it is,
its centroid can be found by loading each of its portions, into
which for purposes of calculation it may be divided, with a weight
proportional to its area, and treating the question as one for the
determination of the C. of G. of a number of isolated weights.
Example 25. — Find the position of the centroid of the Tee section
shown in Fig. 65.
We may consider the section to be made up of two rectangles;
then —
f a 5 SO 24O
Area of flange = 6 X £ = Q sq. ins. = -r— sq. ins.
o o ^4
and the centroid of the flange is at Gj.
CENTROIDS
221
Area of web = 3$ x ~ = -~ sq. ins.
o 04
and the centroid of the web is at G2.
From considerations of symmetry we see that the centroid of the
section must lie on the line G^G^, at the point G, say.
(of length > + , i. e., 2") as a bar loaded with -~-
I O ID 04
Treat G
joe
units at Gj and ~ units at G2.
Let GjG = x', then the upward force at G = 7^ + ~^
= =^ units.
64
,,G,
B
847 —
•344-4'5l6
Q G> I G
I __ i J\ r\f\ /»"
t11^— -* I
'22
FIG. 66. — Centroid of Bridge Rail.
(In the further calculation we may disregard the denominators, since
they are alike.)
Taking moments about
whence
375 X^ = 135X2
x = ^^ = -72.
375 '
Hence the distance of the centroid from the outside of the flange —
'
Example 26. — Determine the position of the centroid of the bridge
rail section shown at (a), Fig. 66.
This example presents rather more difficulty than the one imme-
diately preceding it. The plan of procedure is, for cases such as this,
222 MATHEMATICS FOR ENGINEERS
that adopted in the work on the calculation of weights, viz., we first
treat the section as " solid " and then subtract the part cut away.
Neglecting the small radii at the corners, and treating the section
as " solid," the section has the form shown at (b), Fig. 66.
2 3 *7
The area of AB = -i-X- = 2-52 sq. ins., and its centroid is at G2,
10 4
the intersection of its diagonals
•3 tj
Similarly the area of CD = |x- = 1-313 sq. ins., and its centroid
is at Gj.
For the part cut away (see (c), Fig. 66)
The area of EHM = -Xy!) = -221 sq. in.; and we know from
Part I, p. 130, that its centroid G8 is distant -424 X radius, i.e.,
•424 X -375 or *I59* from EM.
Again, the area of EF = — >X- = -516 sq. in., and its centroid is
I o 4
at G4.
Our problem is thus reduced to that of determining the C. of G.
of four isolated weights, two of which act in the direction opposed to
that of the others, placed as shown at (d), Fig. 66.
Let the centroid of the whole section be at G, distant x from O.
Now the upward forces = the downward forces
and thus RG+'5i6+-22i = 2-52 + 1-313
whence RG = 3-096.
Also, by taking moments about O —
(3-096 X *) + (-516 X -344) + (-221 X -847) = (1-313 X-i88) + (2-52X1-094)
whence x = -855 in.
or the centroid of the section is '855" distant from the outside of the
flange.
Centroids found by Algebraic Integration. — Suppose that
the equation of the bounding curve is given, then the centroid of
the area between the curve, the axis and the bounding ordinates
may be determined by algebraic integration.
We have already seen that —
Ixydx ~lyzdx
x = -. and y = ^—, —
lydx lydx
so that if y is stated as a function of x, xy and y2 may be expressed
in terms of x, and the integration performed according to the
rules given.
The examples here given should be carefully studied, for there
CENTROIDS
223
are many possibilities of error arising due to the incorrect
substitution of limits.
Example 27. — Find the centroid of the area between the curve
y = 2X1'5. the axis of x and the ordinates through x = 2 and x = 5.
The curve is plotted in Fig. 67, and it is seen that the position of
the centroid o^E the area ABCD is required.
Now y= 2x1'5, and thus xy = 2x*xx = 2x*
and 2 = x3.
To find x —
f5
1 xy dx
-f — J * —
\\£dX y
20
c/
/ CenTpoid
/ Vcr»hcal
— '/'***'
I
L
f'ydx
or the centroid ve
i -8 1 units from
boundary.
To find y —
f 2X*dX
\7* )t 15
(F):
?X5|5*_2A ,0
{5*-,*}
5^68 5
JS / ^Cenrnoiol
Ccn^poid
Hopizonrai
A | | | | B
7 50-25
rtical is distant °
the left-hand
i T5
- 1 y*dx
2 J 2 '
23 4 5 J2
FIG. 67.
y — ,5
/ y dx
-X4 / ^3^
2 ^ 7 2
/5 3 /2 6\5
^-^ (V)^
i,,5 (54~2«)
5 6oQ -,
8^50-25
Hence the co-ordinates of the centroid are 3-81, 7-57.
Example 28. — The bending moment curve for a beam fixed at one
end and loaded uniformly over its whole length is a parabola, as
224 MATHEMATICS FOR ENGINEERS
shown in Fig. 68. The vertex is at A and the ordinate at B, viz.,
BC is — ; the loading being w units per foot and I being the span.
We wish to determine the position of the centroid of the figure
ABC so that we may find the moment of the area ABC about AD,
and finally the deflection at A.
From the equation to a parabola, y2 = ^ax, we see that —
Wl2 , 22
lz = 4« . — , whence 40, = — or y2 = — x
^2 w w
i. e., (ND)2 = - AD.
w
The distance of the centroid from AD = y
r xydy
°
j x&y
D
fl
/
FIG. 68.
Area of ABCD = - of surrounding rectangle = X/X —
_ wl3
All this area may be supposed to be concentrated at its centroid,
and hence the moment of ABC about AD = — ^ x - I — -c
04 o
Now the deflection at A = ^ x moment of the bending moment
diagram about the vertical through A
i wl*
~ El X IP
Hence the deflection at A =
^^
oJil
W/3
= 3^=^, where W = total load.
Example 29. — Find the position of the centroid of a quadrant of
a circle of radius Y.
The equation of the circle is xz-\-y2 = r2
hence y —
= 2— *
so that
xy = xVrz—x2.
CENTRE OF GRAVITY 225
Thus x (and consequently y) —
*-x2 dx
0
Try2
The value of the denominator is — , for it is the area of the
4
quadrant. (This integral would be evaluated as shown on p. 149.)
To evaluate the numerator, let u = rz—xz
then du — — -zxdx
du
or xdx = .
2
X — *
fr r du i
Then— I x Vr2-x* dx = I ~— "5
= ~[o-(+'2)*]
i ,
i/J
f •
3
i
3 4r
— y = Z— = X Qr -424^.
4
Centre of Gravity of Irregular Solids. — The methods
given for the determination of the centroids of irregular areas
apply equally well when solids are concerned. For if A is the
area of the cross section of a solid at any point along its length,
distant x, say, from one end, and the length is increased by a
small amount 8x (and if this is small there will be no appreciable
change in the value of A), then the increase in the volume == AS*
or the increase in the weight = pA8x, p being the density.
The moment of this element about the end = pAS* x x
so that x —
fl
ist moments _ J Q
2 weights
/i
pAdx
Axdx
226
MATHEMATICS FOR ENGINEERS
As before, two cases arise, viz., (a) when values of A and x are
given, and (b) when A is denned in terms of x. To deal with these —
In case (a) plot one curve in which A is the ordinate and x is
n
the abscissa and find the area under it ; this is the value of I A.dx.
Jo
Plot a second curve whose ordinates are the products of
corresponding values of A and x and find the area; this is the
value of the numerator, and division of the latter area by the
former gives the value of r. Thus the centroid vertical is found,
and if the solid is symmetrical about the axis of x, this is all that
is required; otherwise the centroid horizontal must be found, the
procedure being exactly that previously described when dealing
with areas in place of volumes.
An example on the application of this method is here worked.
A
6
5
4
5
2
1
0
<
\
\
c
.A » *
• » «^*
AJC
£4
20
16
)2
8
4
O
\
X
x*
„ -*'
~~^~
•s.
\
.
^>
^
^
^
i
^^
/
•<
i
• —
^-~,
y
^^-»
- —
•~-~^
•*^-^_
F
9
JG
C
0 ^ 4 6-8 1O 1£ 14 1(
FIG. 69. — Problem on C.I. Column.
Example 30. — The circumference of a tapering cast iron column,
16 ft. long, at 5 equidistant places is 9-43, 7-92, 6-15, 4-74 and 3-16 ft.
respectively. Find its volume and the distance of its C. of G. from
the larger end.
The areas must first be found from the circumferences.
Now the area of a circle =
47U
So that the table for plotting reads —
x = distance from larger end (ft.)
o
4
8
12
16
A = area of cross section (sq. ft.)
7-09
4-98
3-o
1-78
•79
By plotting these values the curve EF (Fig. 69) is obtained.
CENTRE OF GRAVITY
227
The figure here given is a reproduction of the original drawing to
rather less than half-size, and since the measurements were made on
the original, the results now stated refer to that.
In the original drawing the scales were : i* vertically = 2 sq. ft.,
and i* horizontally = 2 ft., so that i sq. in. of area represented
4 cu. ft. of volume. The area under the curve EF was found, by
means of the planimeter, to be 13-66 sq. ins., and accordingly the
volume = 13-66x4 = 54-64 cu. ft.
The curve BCD results from the plotting of values of Ax as
ordinates, the table for which plotting reads —
X
o
4
8
12
16
Ax
0
I9'9
24
21-4
12-6
The area under this curve was found to be 19-06 sq. ins., which
FIG. 70. — C. of G. of Solid of Revolution.
represented I9~o6x 16 units of moment, since for the plotting of BCD
i* vertically = 8 units of Ax, and i* horizontally = 2 units of x.
area BCDG i6x 19-06
Hence
area BEFG 54-64
For case (b), when A is stated in terms of x, the integration is
entirely algebraic. Thus if A is a function of x, integrate Ax and
also A with regard to x, and divide the former integral by the
latter to determine the value of ~x.
Example 31. — The area of cross section of a rod of uniform density
varies as the cube root of the distance of the section from one end ;
find the distance of the C. of G. from that end, being given that the
area at a distance x from the end = '^/#.
228 MATHEMATICS FOR ENGINEERS
Consider a strip distant x from the stated end and of thickness S.v.
Then, from hypothesis, the area of section — 4- $3/ x, and thus the
volume = area X thickness = 4-5^/^xS^.
Also the mass of the strip = volume X density
and the moment of the strip about the end — mass x distance
2 ist moments of small elements
Hence x = - ...
2 their masses
= 4*
7
or the C. of G. is distant ^ of the length from the given end.
C. of G. of a Solid of Revolution. — Suppose that the curve
BC in Fig. 70 rotates round OX as axis ; and we require to find
the position of the C. of G. of the solid so generated.
Consider a small strip of area MN; its mean height is y and
its width is 8x, so that the volume generated by the revolution
of this is Tiy28x, or the mass = p-n:y28x. The ist moment of this
strip about OY = mass X distance = p-n:y28x X x = pnxy28x.
Thus the total ist moment about OY = /, pnxy28x
*~~^a
^\b
and the total mass = > pny28x
ib fb
I piixy2dx I xy2dx
J_ a J_a
/* ( b
pny2dx I y2dx.
J a J a
As before, the two cases arise, viz. —
(a) When values of x and y are given. For this case make a
table of values of x x y2 and also one of values of y2.
Plot the values of xy2 against those of x and find the area under
the resulting curve
This area = fxy2dx . . . . . . (i)
CENTRE. OF GRAVITY
Plot the values of yz against those of x —
Area of figure so obtained = fyzdx
229
(2)
and
(2)'.
Also we know that y must be zero, for the axis of x is the axis
of rotation; and thus the C. of G. is definitely fixed.
(b) When y is expressed as a function of x. In this case find
both xyz and also yz in terms of x, integrate these functions
algebraically and thence evaluate the quotient.
Example 32. — The curve given by the tabulated values of y and x
revolves about the .ar-axis; find the position of the C. of G. of the
solid thus generated.
X
o
i
2
3
4
y
8
10
21
26-4
25
For the first curve, values of xyz are required, and for the second
curve, values of y2; these values being —
X
o
i
2
3
4
y*
64
IOO
441
696
625
xy2
o
IOO
882
2088
2500
The curve AB (Fig. 71) is obtained by plotting the values of xyz
as ordinates; and the area under this curve is 4323; this being thus
the value of / xy2dx.
J o
By plotting the values of y- as ordinates the curve CD is obtained ;
/i
yzdx = 1699.
o
/4
xyzdx
_o
T*
I y*dx
J o
i. e., the C. of G. is at G, the point (2-55, o).
Example 33. — The curve x — 5V — 2 Vy revolves about the axis of y.
Find the position of the centre of gravity of the solid generated, the
solid being bounded at its ends by the horizontal planes distant i and
5 units respectively from the axis of x.
4323
1699
= 2-55 units
230
MATHEMATICS FOR ENGINEERS
Since the revolution is about the axis of y and not that of x, y must
take the place of x in our formulae and x the place of y; therefore
the limits employed must be those for y.
In Fig. 72 AB is the curve x = $y—2Vy, and we see that it is
2000
5oo_
~S~
FIG. 71.
required to find the height of the centroid above the axis of x of the
solid generated by the curve AB about the axis of y.
Then to find y —
(•5
/ yx2dy
J i
/5
x2dy
i
Now x = 5y—2Vy, and thus x2 = 2
and y#2 — 2$y3— 2oy*-\-4y2.
f5 f5 . 4
Then / y^2ay — I (2^y3—2oy-}-^y2)dy =
y =
= 2454
f5 »j T5
and / x2dy=
J i I i
j T25V3 , 4V2 20X2 |"15
dy = \ ^-~+~ -- ^— y \
L 3 * 5 ->i
= 639
/:
2454
639
= 3'
CENTRE OF GRAVITY
231
Then since the centroid must lie along the axis of y, its position
is definitely fixed at the point G, viz., (o, 3-84).
Example 34. — Find the mass and also the position of the C. of G.
of a bar of uniform cross section a and length I, whose density is
proportional to the cube of the distance from one end.
Let us consider a small length 8x of the bar, distant x from the
end mentioned above; the density of the material here = Kx3, where
K is some constant; hence —
o e 4 6 a o IE • H- 16 IB 20
FIG. 72.
Mass of small element = volume X density = a8x-X K#3 = Kax38x.
Thus the total mass = [ ' Kax3dx = Ka f— V
J o v 4 * "
Ka/4
Also the ist moment of the element about the end —
= mass x distance
= Kax38xxx.
Ka/5
Total ist moment
-71.
Kax*dx =
and if x = distance of C. of G. from the lighter end —
Ka/5
5 L
Example 35. — Find the position of the C. of G. of a triangular
lamina whose density varies as the distance from the apex. (Let the
thickness of the lamina = /.)
232
MATHEMATICS FOR ENGINEERS
Consider a small strip of width 8x, distant x from the apex
(Fig- 73)-
The area of the strip = y>8x, and thus its volume = yt8x.
Now the density «r x or density = Kx
and also, by similar triangles,
Bx
So that the mass of the strip
H
— ytSx x Kx
BKt „
H ^
and the ist moment of the strip about OY —
jr
ri
S.W.S.L.
r
/ -
'\
1 \6x?
1
A- -y
_J_^
/ i
; '\
rE-
*J 7,-
u p
J
? f^*
^U
D
FIG. ;
Hence —
'3-
_ /
x — - 2
/o
(f
FIG. 74
T_ x3dx
ti
BK/
)oE H-,3
(v3\Ji A T~T^
x \ 4 rz
!/•
Centre of Pressure. — If a body is immersed in a liquid, then
the pressure per sq. in. of surface is not uniform over the solid,
for the pressure is proportional to the depth. The point at
CENTRE OF PRESSURE 233
which the total pressure may be supposed to act is known as the
centre of pressure (C. of P.).
To find positions of centres of pressure we are, in effect, finding
centres of gravity of solids whose density is proportional to the
distance from some fixed axis.
The C. of G. found in the example last worked is in reality
the C. of P. of a triangular lamina immersed vertically in a liquid,
with OY as the level of the top of the liquid.
Just as, when discussing the stability of solids in air, we have
supposed the whole mass to be concentrated at the C. of G., so
now, when the solid is immersed in a liquid, the total pressure
may be assumed to act at the one point, viz., the C. of P.
To find the positions of the C. of P. for various sections and
solids we must start from first principles, dealing with the pressure
on small elements, and then summing.
Example 36. — Find the whole pressure on one side of a rectangular
sluice gate of depth 5 ft. and breadth 3 ft., if the upper edge is 10 ft.
below the level of the water (which we shall speak of as the still
water surface level or S.W.S.L). Find also the depth of the centre
of pressure.
Consider a strip of the gate Sx deep and x ft. below S.W.S.L.
(Fig. 74)-
Then the area of the strip =3x8*
and the pressure per sq. ft = K x depth.
Now at a depth of x ft. the pressure per sq. ft. = weight of a
column of water x ft. high and i sq. ft. in section, i. e., wt. of x cu. ft.
of water or 62-4^ Ibs.
Also the pressure is the same in all directions;
and thus the pressure on the strip = 38x^x62-4^
and the moment of the pressure on the strip about S.W.S.L. —
f!5
Hence the total pressure = I i8j-2xdxlbs.
J 10
o (x*\™
= 187-2 - )
\2/10
= 187-2 XJC25 lbs
2
— 11700 lbs. or 5-23 tons.
Again, the total ist moment about S.W.S.L. —
ri5 /x3\
= i8rzx2dx = i87-2(- )
Jio \3/
io
= 62-4x2375.
234
MATHEMATICS FOR ENGINEERS
'-hus the depth of the C. of P. below S.W.S.L. —
= 62-4 X 2375 ft
11700
= 12-65 ft.
Hence C. of P. is at the point P, at a depth of 12-65 ft- below the
surface of the liquid.
The more general investigation for the position of the C. of P.
is given in Chap. X.
Mi
FIG. 75. — Centroids.
Exercises 19. — On the Determination of the Positions of Centroids and
Centres of Gravity.
1. The density of the material of which a right circular cone is
composed varies as the square of the distance from the vertex. Find
the position of the centre of gravity of the cone.
CENTROIDS 235
2. The equidistant half-ordinates of the load water plane of a ship
are as follows, commencing from forward : -6, 2-85, 9-1, 15-54, J8, 18-7,
18-45, 17-6, 15-13 and 6-7 ft. respectively. Find the area of the load
water plane and the longitudinal position of its centroid. The length
of the ship on the load water line is 270 ft.
3. A triangular plate of base 5" and height 8" is immersed in water,
its base being along the S.W.S.L. Find the total pressure on the
plate and the depth of the centre of pressure if the plate is vertical.
4. A vertical retaining wall is 8 ft. wide and 15 ft. deep. Find
the depth of the centre of pressure of the earth on the wall.
5. Draw the quadrant of a circle of 4" radius, and by the double
sum curve method determine the position of its centroid.
6. The portion of the parabola y = 2xz—gx below the x axis
revolves about that axis. Find the volume of the paraboloid so
generated, and the distance of its C. of G. from the axis of y.
7. Find the position of the centroid of the area bounded by the
curve y = 1-7 — 2XZ, the axis of x and the ordinates through x = — i
and x = +4.
8. Reproduce (a), Fig. 75, to scale (full size), and find the position
of the centroid of the section represented, employing the method
outlined in Example 23, p. 216.
9. Draw a segment of a circle of diameter = 6* on a base of 5-92",
and find by the method of Example 23, p. 216, the height of the
centroid above the base. (Take the segment that is less than a
semicircle.)
Find the distance of the centroid from the line AB for the sections
in Nos. 10, ii and 12.
10. Channel Section, (&), Fig. 75.
11. Unequal Angle, (c), Fig. 75.
12. Tee Section, (d), Fig. 75.
13. Make a careful drawing of (a) Fig. 76, which represents the half-
section of the standard form of a stream, line strut for an aeroplane,
taking t as 2", and by the method of Example 23, p. 216, determine
the distance of the centroid from the leading edge.
14. Find the position of the centroid of the pillar shown in Fig. 57,
p. 210, of which further explanation is given in Question 2 on p. 210.
[Deal with the flanges and the body as three separate portions.]
15. One end of a horizontal water main 3 ft. in diameter is closed
by a vertical bulkhead, the centre of the main being 35 ft. below the
level of the water. Find the total pressure on the bulkhead.
16. A semicircular plate is immersed vertically in sea water, its
diameter being along the water surface. Find the total pressure on
the plate if its diameter is 12 ft. and the weight of i cu. ft. of sea
water is 64 Ibs. ; find also the depth of the centre of pressure. [Note. —
The reduction formulje given on p. 178 assist in the evaluation of the
integrals.]
17. The parabola y2 = 6x revolves about the axis of x. Find the
distance from the vertex of the C. of G. of the paraboloid thus
generated, if the diameter of the end of the paraboloid is 18.
236
MATHEMATICS FOR ENGINEERS
(a)
(b)
FIG. 76.
MOMENT OF INERTIA
237
18. The diameter of a spindle at various distances along its length
was measured with the following results —
Distance from end (ins.)
o
i
2
•83
3
4
5
6
7
8
2
Diameter (ins.) .
i'5
I-I2
•85
1-18
i'5
1-78
1-96
Find the distance of the C. of G. from the smaller end.
19. Find, by means of the double sum curve method, the distance
from AB of the centroid of the rail section shown at (a), Fig. 75.
20. An aluminium right circular cone is of height 7 ins. and the
diameter of its base is 10 ins. Find (a) its mass, the density of
aluminium being -093 Ib. per cu. in. ; (6) the height of its centroid
above the base.
21. Use the double sum curve method to find the distance from
AB of the centroid of the area shown at (6), Fig. 76.
22. A segment of a parabola is of height h and stands on a base b
Find the height of the centroid above the base.
23. A triangular plate of height h is immersed in water, its vertex
being at the water surface, and its base being horizontal. Find the
depth of the centre of pressure of the plate.
Moment of Inertia. — The product of a mass into the square
of its distance from some fixed point or axis is called its second
moment about that point or axis; and for a number of masses
the sum of their respective second moments becomes the second
moment, or moment of inertia of the system. When the number
of masses is infinite, i. e., when they merge into one mass, the
limiting value of the sum of the second moments is spoken of as
the moment of inertia of the body.
The moment of inertia of a section or body determines to a
large extent the strength of the section or body to resist certain
strains ; the symbol I, which always stands for moment of inertia,
occurs in numerous engineering formulae ; also when dealing with
the formulae of angular movement the mass is replaced by I, and
so on, so that it is extremely important that one should be able
to calculate values of I for various sections or solids.
A few examples will emphasise the frequent recurrence of the
letter I. Consider first the case of a loaded beam : —
Let the figure (Fig. 77) represent the section of a beam loaded
in any way. Then it is customary to make the following
assumptions —
(a) There is to be no resultant stress over the section, i. e., the
sum of the tensions = the sum of the compressions.
238 MATHEMATICS FOR ENGINEERS
(b) That the stress varies as the strain, and that the Young's
modulus for the material is the same for tension as for compression.
(c) That the original radius of curvature of the beam is exceed-
ingly great compared with the dimensions of the cross section of
the beam.
The surface of the beam which is neither compressed nor
stretched is spoken of as the neutral surface, and the line in which
this cuts any cross section of the beam is known as the neutral
axis.
Referring to Fig. 77, let NN be the neutral axis, and let o- be
FIG. 77.
the stress at unit distance from NN, i. e., a-y = the stress at a
distance y from NN.
Thus the stress at y on a section of breadth b and depth 8y = a-y,
and the force = stress X area = b8y X <ry.
Now the forces on one side of NN must balance those on the
other (by hypothesis).
rr,
bdya-y = o.
but I 1 a-bdy x y = total ist moment of the forces
and the line about which this is zero must pass through the centroid
of the section ; hence the line NN passes through the centroid.
The tensile and compressive forces form a couple, the moment
of which —
= 2 force x distance = ;x \ b8y<ry x y
MOMENT OF INERTIA 239
i. e., in the limit the moment of resistance of the internal forces
= o- / bdyxy2, i. e., a- /area x (distance)2
J ~Y2
i. e., a- (2nd moment of section about NN)
= o-I.
If M is the bending moment at the section, i. e., the moment
of the external forces, it must be exactly balanced by the moment
of the internal forces, so that M = oT.
Also if /j = maximum tensile stress and = o-Yj
/2 = maximum compressive stress and = o-Y2
A ft M
then o- = ±± = ±f- = T
*i *2 L
M /
or, in general, T ^ v~
Hence, in considering the strength of a beam to resist bending,
it is necessary to know the moment of inertia of its section;
knowing this and the bending moment, we can calculate the
maximum skin stress.
As a further illustration of the importance of I in engineering
formulae let us deal with the following case : If a magnet is allowed
to swing in a uniform field, the time T of a complete oscillation is
given by —
where I = moment of inertia of the magnet
M = magnetic moment of the magnet
H = strength of the uniform field in which the
magnet swings.
In this case the I of a cuboid is required; and it will be seen
that no mention of the mass is made in this fortnula. Actually
the I takes account not only of the mass, but also of its disposi-
tion, the latter being a most important factor in all questions of
angular movement. Thus for a mass of i Ib. swinging at the end
of an arm of 10 ft. the energy would be io2, i. e., 100 times that
of the same mass placed at a radius of i ft. only, although the
angular velocities in the two cases were the same.
The reason for the presence of I in formulae concerning the
energy of rotation will be better understood if the next Example
is carefully studied.
240
MATHEMATICS FOR ENGINEERS
Example 37. — A disc revolves at n revs, per sec. ; find an expression
for its energy of rotation, or its kinetic energy.
If the total mass = ra, let a small element 8m of mass be considered,
distant r from the axis of rotation (Fig. 78).
Now the linear velocity at the rim = V = 2?rwR
and the angular velocity — &> = number of radians per sec.
then
or
Ra, = 2TCWR = V
V
thus co is constant throughout, whilst V depends
on the radius.
Kinetic Energy of mass 8m
_ massx (veloc.)2 _ 8mxv2
FIG. 78.
Hence the total K.E. of the disc =
= rz8m.
-r2dm
_ coa fn
2£ Jo
massx (distance)2
= w Xl for disc.
2£
Thus the K.E. = — Io>2. Comparing this formula with the cor-
responding one for linear motion, viz., K.E. = — mvz, we see that
~6
when changing from linear to angular movement, I takes the place of
m and <a the place of v.
Suppose that the average velocity — vt = r^
then — my,2 = Io>2
2g
i. e., mr
or
Hence I is of the nature of mass x (distance) 2, so that if the whole
mass were concentrated at the distance rt from the axis, the K.E. of
the system would be unaltered.
Hence the distance rt (which is usually denoted by k) is referred
to as the swing or spin radius, or radius of gyration, i. e., it is the
effective radius as regards all questions of rotation.
MOMENT OF INERTIA 241
[Note that k is not the arithmetic mean of the various radii, but
the R.M.S. value for—
h = J * (radius)*
number considered J
number considered
In general, I can be written as mk2 (if dealing with a mass) or
Ak2 (if concerned with an area).
Method of Determination of the Value of I for any
Section. — Whilst it is found desirable to commit to memory the
values of I for the simpler sections, it is not wise to trust entirely
to this plan. It is a far better policy to understand thoroughly
the meaning of the term "moment of inertia," and to derive its
value for any section or solid by working directly from first
principles.
Thus, knowing that the moment of inertia is obtained by
summing up a series of second moments, we divide the area or
mass into a number of very small elements, find the area or mass
of each of these and multiply each area or mass by the square
of its distance from the axis or point about which moments are
required ; the sum of all such products being the value of I.
If the length of the swing radius is required, it can be deter-
mined from the relation I = Ak2 (for an area) or I = Mk2 (for
a solid) ; the area or mass being obtained by the summation of
the areas or the masses of the separate elements.
T« , /2 second moments of elements
Thus k = \i= — , , — — .
\ 2 areas or masses of elements
Confusion often arises over the units in which I is measured;
and to avoid this it is well to think of I in the form Ak2 or M&2,
when it is observed that I is of the nature area x (length)2, *'. e.,
(inches)2 x (inches)2 or (inches)4 for a section, and massx (length)2
or Ibs. X (inches)2 for a solid.
The moment of inertia must always be expressed with regard
to some particular axis; and it is frequently necessary to change
from one axis to another. To assist in this change of axis the
following rules are necessary : —
The Parallel Axis Theorem. — By means of this theorem, if
I is known about an axis through the C. of G., the I about an
axis parallel to the first can be deduced.
In Fig. 79 NN is the neutral axis of the section; and the
moment of inertia is required about AB, i. e. IAB is required.
R
242 MATHEMATICS FOR ENGINEERS
Dealing with the strip indicated—
IAB of the strip = pb8y x y2.
Hence the total IAB = pfbdyxy2
'= Pfbdyx(Y-d)2
= pfbdy X Y2+P/My X d2-2PfbdyYd.
Now fpbdy X Y2 = the total INN
and fpbdy X dz = d2Jpbdy = d2 X total mass = md2
also 2dfpbdyxY — 2d x total ist moment about NN
= 2d X o (for the moments on the strips on
one side' of NN balance those on the
other)
= o.
Hence — IAB = INti-{-md2
i. e., to find the moment of inertia about any axis, find the moment of
&/
FIG. 79.
inertia about an axis through the G. of G. parallel to the axis given, and
to this add the product of the mass into the square of the distance
between the axes.
e. g., if INN = 47, mass = 12-4 and d (between AB and NN=2'3)
then IAB=INN+rf=47+(i2-4X2-32)
= 47+657 = II27-
Since IAB = Ij
then w&AB = n
or
k"AB —
and this relation is represented by Fig. 79, which suggests a graphic
method of finding &AB when £NN is known.
MOMENT OF INERTIA
243
Theorem of Perpendicular Axes. — We require to find I
about an axis perpendicular to the plane of the paper and passing
through 0 ; such being spoken of as a polar second moment.
To distinguish between the moment about an axis perpendicular
to the plane of the paper and that about an axis in the paper,
we shall adopt the notation I0 for the former and Iox or IOY, as the
case may be, for the latter.
To find Io : —
Consider a small element of mass 8m at P (Fig. 80).
Then Iox of this element = 8m x y2, IOY — 8mx x2,
r2 = x2+y2
and I0 = 8m xr2.
8m. r2 = 8m.x2+8m.y2
fdm . r2 = fdm . x2+fdm . y2
total I0 = total IOY+ total Iox
I0 =
FIG. 80.
Now
hence
and
*'. e.,
or
so that if the moments of inertia about two
perpendicular axes in the area are known,
the sum o! these is the moment of inertia
about an axis perpendicular to the area and
through the point of intersection of these axes.
In special cases for which Iox = Ioy
then I0 = 2lox
To find the Relation between the Moment of Inertia
about a Point in a Solid Body and the Moments of Inertia
about three mutually Perpendicular Axes meeting in that
Point.
Thus, referring to Fig. 81, it is desired to connect Io with Iox>
IOY and Ioz.
Consider a small element of the mass 8m placed at the point P.
Then if PS = x PT = y PM = z OP = r
(ON)2+(NM)2+(PM)2 = (OP)2, and ON = PS, NM = PT
*. e., x2-\-y2-\-z2 = r2.
Now Iox of the element = Sw(PN)2 = 8m(z2+y2)
and in like manner IOY = 8m(x2+z2) and Ioz = 8m(y2+x2)
also I0 = Sw(QP)2 = 8mr2.
Thus I0 = Smr2 = Sw(*2+y2+*2)
= 8m
/
244 MATHEMATICS FOR ENGINEERS
And for the total mass —
or
total I0 = -(
We may now apply the principles already enunciated to the
determination of the moments of inertia of various sections and
FIG. 81.
solids ; and we take as our first example the case of a rectangular
section.
Example 38. — To find the moments of inertia of a rectangle about
various axes.
(a) To find INN (Fig. 82), NN being the neutral axis.
Dealing with the small strip, of thickness Sx —
INN of strip — bSx x x2 i. e., area X (distance)2
h h
Hence the total INN = P bxzdx = b(-3}° = —
J * \3/ h I2
~
A
N,
D
M
c
N
|
»
ai
I
B
.lb-~
-_i
IN,
FIG. 82.
= area X —
12
but
where A is the area of the section
AZ_ _ A/fc \2
A^ =
12
or
MOMENT OF INERTIA
245
By symmetry it will be seen that
(6) To find IAB.
_ Ab2
1N1N1 ~: -^
and the distance between AB and
hence IAB =
bz
= A X — .
= b
2
= A —
= A62
3 '
IAB is larger than IN^. as would be expected, for the effective
radius must be greater if the plate swings about AB than if it swings
about N^NV
I c« !
rTT" ; b 1 s*1 i ...
NLJ-— ._ P-J&. ^ZJfrpJNi
n
*4
I
$
N •;
In like manner —
if
FIG. 83.
Ahz
LAD
The rule for the moment of inertia of a rectangle is required very
frequently, since many sections can, be broken up into rectangles.
Example 39. — To find INN of the Tee section shown in Fig. 83.
The neutral axis NN is distant 1-03* from AB (cf. Example 25, p. 220).
Dealing with the flange —
i i /S\3
INiNj = — bh3 = ^x6x(J] : : *122 in>
also the distance GZG = •72".
—- Hence by the parallel axis theorem —
246 MATHEMATICS FOR ENGINEERS
INN of the flange = INiNj + [Ax (GXG)2] (A being the area of the
flange)
5,,
2) = -I22+I-94
= 2-06 ins.4
For the web iNrfl> = ^6A» = -L X ^ )* x |- = 2 ins.4
and also G2G = 1-28".
Hence INN of the web = IN2N2+ Ax x (G2G)2 (Ax is area of the web)
= 2 + (fxfxi-282)
= 5'45 ins.4
Hence the total INN of the section = 2-06+5-45 = 7-51 ins.4
Example 40. — Find the polar 2nd moment of a circular disc of
radius R; and also the moment of inertia about a diameter.
Consider a ring of width 8r, distant r from the centre (Fig. 84).
Then Io of the annul us = mass (or area) x (distance)2
Hence
the total I
Now
and
FIG. 84.
To find the respective swing radii —
A£Q = T0 = • ~^~
TUR4
i. e.,
ox
(Cf. with the R.M.S.l
•D
or kQ — , i. e., -TofR. -! value of a sine function V
V* ( of amplitude R. J
Also
MOMENT OF INERTIA
247
ox
4X:rR2
= - = -sR.
R2
4
To find the swing radius about a tangent —
(distance)2 (oxtoTT) = R2
IYT = Iox+AR2
hence
or
Example 41. — To find the moment of inertia of a right circular
cylinder of length h and radius R, about various axes.
(a) About the axis of the cylinder.
(b) About an axis through the C. of G. perpendicular to the axis
of the cylinder.
(c) About an axis parallel to that in (b), but through one end.
(a) The 2nd moment about the axis of the thin cylinder of
length Sx (Fig. 85)—
R2
= mass X — from Example 40
R2
= pnR28x x — p being the density of the material .
Hence the total 2nd moment about the axis —
.R2
= r
J 0
dx =
R2
2
R2
= m -
•2
where m = the mass of the cylinder.
248
MATHEMATICS FOR ENGINEERS
(b) The 2nd moment of the strip about AA, which is parallel
to NN—
R2
= mass x - (see Example 40)
R2
Hence INN of the strip = IAA of the strip + (its massx*2)
since AA is an axis through the C. of G. of the strip.
Thus INvr of the strip = Sx + uTiR2x28x
4
h h
f2 oTtR4 /"2
and thus the total INN = j — dx-\- I pnR2x2dx
J _h 4 J _ h
N
(c) To find IFF, FF being parallel to AA and NN.
The distance between FF and NN = -
2
Also it has just been proved that IN1j = m ( —
Hence
T /R2 , h*
!„„ = ml —
\4 12
/R2 , h2\
= m(- + --).
Example 42. — Find an expression for the moment of inertia of a
large pulley wheel of outside radius R and thickness of rim /. Neglect
the arms or spokes of the wheel.
Let Y — inside radius of wheel, i. e., r = R— t.
Then, using the result of Example 40, p. 246, we know that the
TD4
moment of inertia of the wheel as solid =
; from this must
MOMENT OF INERTIA 249
7W4
be subtracted the moment of inertia of a disc of radius r, viz., — Xpb
(p being the density of the material and b the breadth of the rim
along the face).
, , /T,.
Hence IQ - Pb-~Pb = -~ (R*-y4
= (R2+r>) (R2-*
. . . „,
where M is the mass of the wheel.
M M
Writing R-/ forn!0 = ^ (R2+R2+/2-2R/) = ™ (2R2-2R*+*2).
From (i) it will be seen that in order to get Io as large as possible,
R and r must be very nearly equal, i. e., t must be very small compared
with R. Thus for an approximation 22 may be neglected in the
M
expression for IQ, so that IQ = -xaR (R— t) = MR(R— t).
/R2+y2\ 2 /R2+y2\
Referring once again to (i), Io = M ( — •*-—)* i- &-, M&o = M ( - ! — J
or kQ = - and ko = -jojVRt+r2. As an approximation for
this the rule kQ = - (R-\-r) is often used; kQ being thus taken as the
average radius.
Moment of Inertia of Compound Vibrators. — To find the
modulus of rigidity of a sample of wire by the method of torsional
oscillations, various forms of vibrators may be used. In the
calculations which follow the experiments, the moment of inertia
of the vibrator occurs, so that it is necessary to understand how
to obtain this. To illustrate by an example of one form of
compound vibrator, suppose that the I about an axis through
the C. of G. of the one shown in Fig. 86 is required.
Let m{ = mass of AB, rl be its radius and ^ its length
m2 = mass of C and also of D, rz be its radius and /2 its length.
fr 2 / 2\
Then INN of AB = wa( — +-1 ) (from Example 41, p. 248)
c r irzz , ^22\ rf (f°r tne inner radius\
and IofC^m-1 = ^
250 MATHEMATICS FOR ENGINEERS
by the parallel^
- mz\^ -rI2y-rw2t- mz ^ \
This is also the L
•\4
of D.
2
4 V axis theorem. )
LNN
total INN =
Maxwell's needle is a very convenient form of compound vibrator,
and is utilised to determine the modulus of rigidity of the sample
of wire by which it is supported. It consists essentially of a tube
along which weights may be moved from one position to another,
the movement being a definite amount.
Referring to Fig. 87 —
m^ = the mass of each of the movable weights
w2 = the mass of each of the fixed weights.
FIG. 87.
FIG. 88.
Then the time of torsional oscillations is measured when the
movable weights are placed as shown, and again when they are
moved to the centre; and it can be proved that the modulus of
rigidity depends upon the difference between the moments of
inertia under the two sets of conditions.
Thus, since a mass mt is shifted from the position AB to the
position NA, the only difference in the moments of inertia is that
due to the changing of the C. of G. of a mass (m^— m2) from a
distance fa from the axis of oscillation to Ja; for INN of m.2 is
unaltered.
q
r
i6
Hence the change of I \ , T
considering one mass onlyJ " l 2 ~~ ^•mi~i"'^\-L^r ^
(j \
a2)
= (wx— m2)a2.
Example 43. — Find the moment of inertia of a sphere of radius R
about its diameter.
MOMENT OF INERTIA 251
Consider the thin disc (Fig. 88) of radius y, and thickness 8x.
I of the strip about a diameter parallel to OY —
— y*p8x (cf. Example 40, p. 246).
Hence I of the strip about OY (distant x from the diameter
considered)
= T5"1
Now y* = R2— x*.
Thus IOY of disc = TTP J ""^ ~^v" "^ ^ 18*
r I 4 ' /
and hence TOY of sphere = — I (R4— ;
= 27rp[R(R* -
4 J o
= R^ I
4 L 5 i
27uo i6R5 8
= — - X — — =
4 15 !
5
2 ( m being the mass\
- 7/fr /\ ~ JLV \ tii i
5 \ of the sphere. /
and
Determination of 1st and 2nd Moments of Sections
by means of a Graphic Construction and the Use of a
Planimeter.
The graphic construction now to be described is extremely
simple to understand, and has the additional merit of being
utilised to give 3rd, 4th and higher moments if desired.
It being required to find the ist and 2nd moments about MM
of the rail section shown in Fig. 89, and also the position of the
neutral axis, the procedure is as follows : —
Construction. — Divide the half -area into a number of strips by
means of horizontal lines; the half-area only being treated, since
the section is symmetrical.
At a convenient distance h from MM draw MXMX parallel to
MM. From P, the end of one of the horizontals, draw PR per-
pendicular to MM, and from P1, the other end of the same horizontal,
drop PKR1 perpendicular to MXMX ; join RXR and note Q, its point
of intersection with P1?. Repeat the process for all the other
horizontals (of which only three are shown in the diagram) and
252
MATHEMATICS FOR ENGINEERS
join up all the points like Q, thus obtaining the curve CQLS, which
is termed the ist moment curve.
To obtain the 2nd moment curve treat the area CPKXSLQ
in the same way as the original area was treated, i. e., drop QR"
perpendicular to M.1M.l and join RR"; join up all points like Q1
and the 2nd moment curve is obtained.
Calculation. — Find by the planimeter the areas of the original
M
M
I^Momenf
Curve
M,
FIG. 89. — Moments of Sections by Graphic Construction.
half-section, CPKXSLQ and CPKXTWQ1 ; call these AC, Ax and
A 2 respectively.
Then ist moment of the section about MM = 2Xhh.^
(for Aj is for the half-section only).
Distance of the centroid of the section from MM = -r~
2nd moment of section about MM =
i. e., (swing radius MM)2 = -^ -
and by the theorem of parallel axes, I can be found about NN.
MOMENT OF INERTIA 253
In this case the actual results are as follows : —
h = 3 ins. A0 = i-n sq. ins. Ax = -573 sq. in. A2 = -39 sq. in.
Hence h = 3 X^73 = 1.55 ins.
i
ist moment of section about MM = 2 X 3 X -573 = 3-44 ins.3.
2nd moment of section about MM = 2X32X -39 = 7-02 ins.4.
Swing radius about MM = \ — — = 1-78 ins.
\ 2-22 — ' -
N.B. — To distinguish which area is to be read off by the
planimeter the following rule should be observed : Read the area
between the ist or 2nd moment curve, as the case may be, and the
side of the original contour from which we dropped perpendiculars
on the line about which we required moments.
Proof. — Consider P1? as the centre line of a thin strip (such
as the one indicated). Then the area of the strip = P^xS*, and
ist moment about MM = PXP x 8x X RP.
From the similar triangles RPQ and RJR1
RP_ J*l h
QP ~ JR1 ~~ PP1
whence RP x PP1 = h X QP
and RPxPPxxS* = AxQPxS*
i. e., ist moment of the strip about MM —
= h x the area of which QP is the centre line.
Then, by summing —
Total ist moment of the half-area about MM —
= h X the area between ist moment curve and right-hand
boundary of section
= h\1.
Again, the 2nd moment of the strip about MM = area x (distance)2
= PP1xS*x(RP)2
and 1
RJ ~~ JR" ~ PQ
RP PQ1 . or> hPO1
• _ - _ ^_ » a T?P - _ S—
h ' ~ PQ PQ '
Hence the 2nd moment of the strip about MM —
= P1PxRPxRPx8*
PO1
^ Sx = h*xPQlx8x
= h2 X area of which PQ1 is the centre line.
254 MATHEMATICS FOR ENGINEERS
And the total 2nd moment of the half-area about MM —
= A2 x area between the 2nd moment curve and the right-
hand boundary of the section.
Exercise 20.— On Moment of Inertia.
1. Find the swing radius about the lighter end of a rectangular
rod of uniform section and breadth and length I, for which the density
is proportional to the square root of the distance from that end.
2. The swing radius of a connecting rod about its centre of suspen-
sion was found to be 35-8 ins., and the distance of the C. of G. from
the point of suspension was 31-43 ins. Find the swing radius about
the neutral axis.
If the connecting rod weighed 86-5 Ibs., find its moment of inertia
about the neutral axis.
FIG. 91.
3. A circular disc, 7" diameter, has a circular hole through it, of
diameter 3", the centre of the hole being \" distant from the centre
of the disc. Find the swing radius of the disc about an axis through
its centre of gravity, perpendicular to the face of the disc.
4. Find the moment of inertia of a rectangle (5" by 3") about a
diagonal as axis.
5. Find the swing radius of a triangular plate (of height h) —
(a) When swinging about its base.
(b) When swinging about an axis through the vertex, parallel to
the base.
6. By dividing into strips, by lines parallel to AB, find the moment
of inertia, about AB, and also the swing radius, of the section shown
at (a), Fig. 75, p. 234.
7. Find the radius of gyration about the axis of rotation, of the
MOMENT OF INERTIA 255
rim of a flywheel, of outside diameter 5' 2", the radial thickness of
the rim being 4".
Find the moment of inertia about the neutral axis of the sections
in Nos. 8, 9 and 10.
8. Channel Section, (&), Fig. 75.
9. Unequal Angle, (c), Fig. 75.
10. Tee Section, (d), Fig. 75.
11. Find the swing radius, about the axis, of a paraboloid, the
diameter of the bounding plane, which is perpendicular to the axis,
being d.
12. The flexural rigidity of a beam is measured by the product of
the Young's Modulus E for the material into the moment of inertia
of the section. Compare the flexural rigidity of a beam of square
section with that of one of the same material but of circular section,
the span and weight of the two beams being alike.
13. A cylinder 6" long and of i \" diameter is suspended horizontally
by means of a long wire attached to a hook, and the wire is then
twisted to give an oscillatory movement to the cylinder. Find the
moment of inertia of the cylinder about the hook.
14. Determine the moment of inertia and also the swing radius
about AB of the rectangular section shown at (a). Fig. 90.
15. Calculate the moment of inertia and also the swing radius of
the box section shown at (b), Fig. 90, both about NN and about AB.
16. Find the position of the neutral axis of the section shown at
(c), Fig. 91, and then calculate the moment of inertia and also the
swing radius about this axis.
17. Determine the swing radius of the -section shown at (d), Fig. 91,
about the axis NN.
18. The moment of inertia of the pair of driving wheels of a
locomotive connected by a crank axle was found by calculation to
be 34133 Ibo. ft.2. If the total weight of the two wheels and the axle
was 8473 Ibs., and the diameter of the driving wheels was 6 ft. i in.,
A2
find the swing radius of the wheel and also the ratio -^, where r is
the radius of the wheel.
19. Find the swing radius about the axis of a right circular cone
of uniform density, the radius of the base being 5 ins.
20. Employing the method explained on p. 251, determine (a) the
ist moment about AB, (b) the 2nd moment about AB, (c) the
distance of the centroid from AB, and (d) the swing radius about AB,
of the area shown at (b) Fig. 76, p. 236.
21. A steel wire, -15 in. in diameter, hangs vertically; its upper
end is clamped, and its lower end is secured to the centre of a
horizontal disc of steel, which is 6 in. in diameter and g in. thick.
If the length of the wire is 3 ft., and if C, the modulus of transverse
elasticity of the steel, has the value 12,540,000 Ibs. per sq. in., find
the time of a torsional oscillation of the wire, from the formula —
* = 402-5
256 MATHEMATICS FOR ENGINEERS
where I = moment of inertia of the disc about the axis of suspension
in Ibs. ins.2, / = length of wire in feet, d = diameter of wire in inches.
22. An anchor ring is generated by the revolution of a circle of
radius r about an axis distant R from the centre of the circle. Find
the moment of inertia of the ring about this axis. (Hint.— Commence
with the polar moment, i. e., the moment about the given axis, of an
annulus made by a section at right angles to this axis, finding an
expression for the inner and outer radii of the annulus in terms of
the distance from the central annulus, and then sum up.)
23. Find the swing radius about the major axis of the ellipse
whose equation is —
CHAPTER VIII
POLAR CO-ORDINATES
Polar Co-ordinates. — A point on a plane may be fixed by
its distances from two fixed axes, or by its distance along a line
which makes a definite angle with some fixed axis. In the former
case we are concerned with rectangular co-ordinates and the point
is written as the point (x, y) ; whilst in the latter case the co-
ordinates are polar and the point is denoted by (r, ff), r being the
length along the ray inclined at an angle 6 to the fixed axis.
It is really immaterial as to what line is taken as the fixed axis :
in many cases the horizontal axis is taken, but in order to agree
with the convention adopted for the measurement of angles (see
Part I, Chapter VI) we shah1 here
consider the N. and S. line, i. e., a
vertical line, as the starting axis and
regard all angles as positive when
measured in a right-handed direction
from that axis. A point is next fixed
on that line from which all the rays
or radii vectors originate, and this
point is spoken of as the pole for the
system : thus the reason for the term
polar is seen.
To illustrate this -method of plot-
ting, let us refer to Fig. 92. Taking
OY as the starting axis and O as the
pole, the point (2, 35°) is obtained by
drawing a line making 35° with OY
and then stepping off a distance OP along it to represent 2 units,
i. e., r=2 and 0=35°. In like manner Q is the point (17, 289°) ;
whilst R is the point (2-4, —20°).
One advantage of this method of plotting is that it is not
necessary to classify into quadrants and to remember the arrange-
s 257
FIG. 92.
258
MATHEMATICS FOR ENGINEERS
ment of the algebraic signs ; all lengths measured outwards from
the pole being reckoned as positive.
Example i . — The following table gives the candle power of an arc
lamp for various positions below the lamp : plot the polar diagram.
Angle below horizontal . .
o
10°
20°
3°°
40°
5°°
60°
70°
80°
90°
1800
800
600
480
In reality we have to plot a number of polar co-ordinates, the
lengths representing the values of the candle power ; but since the
horizontal axis is specified, we shall take that as the main axis. Draw
rays making 10°, 20°, 30°, etc. (Fig. 93), with the horizontal axis, and
along these lines set off distances to represent the respective candle
powers, always measuring outwards from the centre. Join the ends
of the rays and the polar diagram is completed.
The Archimedean spiral and the logarithmic or equiangular
spiral, important in connection
-IOOO ^ with the forms of cams and gear
wheels respectively, may be easily
plotted from their polar equations.
Thus the equation to the Ar-
chimedean spiral is r=aO, and the
equation to the equiangular spiral
is r=aebe; indicating that in the
former case the rays, for equal
angular intervals, are the con-
secutive terms in an arithmetic progression, whilst in the latter case
the rays are in geometric progression.
To illustrate the forms of these curves by taking numerical
examples : —
Example 2. — Plot the Archimedean spiral ^=-573^, showing one
convolution.
FIG. 93. — Candle Power of Arc
Lamp.
In the equation d must be in radians, but to simplify the plotting we
can transform the equation so that values of a (in degrees) may
replace 6 (radians).
Thus — r = -5736 = . (degrees) = -oia.
O / O
POLAR CO-ORDINATES
Then the table for the plotting reads : —
259
a
o
3°
60
9°
120
150
180
210
240
270
300
33°
360
'
•6
2-4
2-7
3'°
3*3
3-6
and the plotting is shown in Fig. 94.
aro
izo"
24O
1.50*
eio°
FIG. 94.
Example 3. — Plot one convolution of the equiangular spiral
•25a
In the log form log r = log -5 + -004360 log e
= T-6990+ (-00436 X -4343 X a)
= 1-6990-!- -ooi894a
and thus the table of values reads : —
a. . .
o
3°
60
90
120
150
1 80
210
240
270
3°°
33°
36o
•00189401
0
•0568
•1136
•1705
•2273
•2841
•3409
'3977
•4546
•5114
•5682
•625
•6818
log r .
1-6990
7-7558
1-8126
f-8695
1-9263
1-9831
•0399
•0967
•1536
•2104
•2672
•324
•3808
r . . .
•5
•5699
•6495
•7407
•8439
•9618
1-096
1-249
1-424
1-624
1-85
2-109
2-403
26o
MATHEMATICS FOR ENGINEERS
The curve is drawn in Fig. 95.
It will be seen that the ratio of
second ray _ -5699 _
first ray -5
third ray = -6495 _ ^
second ray -5699
so that this spiral might alternatively have been defined as one for
which the rays at equiangular intervals of 30° form a geometric pro-
gression in which the common ratio is 1-14, the first ray being '5".
Comparing the given equation r — '$e'259 with that connecting
the tensions at the ends of a belt passing round a pulley, viz.,
T = te*8, we observe that the forms
are identical, or in other words the
equiangular spiral might be used to
demonstrate the growth of the ten-
sion as the belt continuously em-
braces more of the pulley.
Selecting any point P on the
spiral, and drawing the tangent PT
there and also the ray OP which
makes an angle <£ with the tangent,
it is found that cot <£ = -25 = co-
efficient of 6 in the original equation.
This relation would hold wherever
the point P was taken on the spiral,
so that the angle between the ray and the curve is constant : and
thus the spiral is called " equiangular."
If cot <£ = i, <£ = 45° and r = ae6, or taking a = i, r = e6 and
loger = 6. Thus a spiral could be constructed in which the angles
(in radians) would be the values of the logs of the rays : this
spiral, however, is extremely tedious to draw, and its value consists
merely in its geometric demonstration of the relationship between
the natural logarithms and their numbers.
Connection between Rectangular and Polar Co-
ordinates.— Let P be a given point, with rectangular co-ordinates
x and y and with polar co-ordinates r and 0.
Then referring to Fig. 96 —
ON y
OP ,
180°
FIG. 95.
so that
y = r cos 0
and
so that
and also
POLAR CO-ORDINATES
OM x
26l
x
y
x = r sin
r sin 0
r cos 0
= tan 0.
Use of Polar Co-ordinates for the Determination of
Areas. — Polar co-ordinates may be usefully employed to find areas
of certain figures.
It is stated in the previous work on mensuration that —
Area of sector of circle =
where 0 = angle of the sector in radians.
N
P
V
X
FIG. 96.
FIG. 97.
Let P and Q (Fig. 97) be the two points (r, 6} and (r+8r, 0+ SO)
and close to one another.
Then, since r and r+&r differ very slightly
Area POQ =
and the total area AOB =
or
re?
±rzd0
J el
approximately
exactly.
For the evaluation of this integral the working may be either
graphic or algebraic, according to the manner in which the relation
between r and 0 is stated.
As a simple illustration we may take the case of a circle of
radius a. The area of the circle was found at an earlier stage
(see p. 225) by evaluating fydx, i. e., by expressing the integral in
terms of the rectangular co-ordinates. To evaluate the integral,
however, it was found necessary to make the substitution x = a sin 6,
the change thus being from rectangular to polar co-ordinates.
262 MATHEMATICS FOR ENGINEERS
Evidently the rotating ray is constant in length and equal to a,
the radius of the circle, and the limits to 6 are o and 2ir, if the full
area is required ; hence —
fZir [2*
Area of circle = I \d*d6 = Ja2 1 d& = Aa2 . 2ir
Jo Jo
= TO?.
Example 4. — Find the area of the cardioid given by the equation
Y = a (i+cos &], 6 ranging from.o to ZTT.
In this case r is of variable length, but there is a definite connection
between r and 6, so that the integration is algebraic.
and r2 = a2 (i-f-cos <9)2
= a2 (i + J cos 2(9+J+2 cos 0)
= a2 (1-5 + \ cos 20+2 cos 0).
j'Zn-
Hence area = j \vz d6
COS 20 + 2 COS0) dd
2 / \ £TT
= — ( i'50+J sin 20+2 sin 0 j
2 2
The Rousseau Diagram. — The use of the Rousseau diagram
simplifies the determination of the mean spherical candle power of
a lamp.
The candle power of the lamp varies according to the direction
in which the illumination is directed (cf . Fig. 93) ; in the case there
discussed, however, we considered the illumination in one plane
only. If we imagine the polar curve to revolve round the vertical
axis we see that a surface is obtained by means of which the
illumination in any direction can be measured. The mean of all
these candle powers is spoken of as the mean spherical candle
power of the lamp. If the arc is placed at the centre of a spheri-
cal enclosure, of radius R, then, if IM is the mean spherical candle
power (M.S.C.P.) of the lamp, the total illumination is expressed
by 47rR2lM : this total might be arrived at, however, by summing
the products of the candle power in any direction into the area of
the zone over which this intensity is spread; and putting this
statement into the form of an equation,
47rR2IM = 2IA,
where I is the intensity on a zone of surface area A.
POLAR CO-ORDINATES
263
To find the M.S.C.P. proceed as follows: — Suppose that the
lamp is at O (Fig. 98). With centre O and any convenient radius
R describe a semicircle ; also let the polar diagram be as shown
(the curve OPQMC) . The greatest candle power is that given by
OC ; draw a horizontal through N, the point in which the line OC
meets the circumference of the semicircle, and make ab = OC.
Through a and b draw verticals and through A and B draw
horizontals, thus obtaining the rectangle DE ; draw a number of
rays, OP, OQ, OS, etc., and also horizontals through the points
p, q, s, etc., marking along these lines distances equal to OP, OQ,
OS, etc., working from DF as base. By joining up the points so
obtained the curve FL6D is obtained, known as the Rousseau
FIG.
curve ; then the mean height of this curve (which can readily be
obtained by means of a planimeter) gives the M.S.C.P. of the lamp.
Proof of this Construction. — Let IM = M.S.C.P. of the lamp
2 area of zonexC.P.
then
4irR*.
Consider the zone generated by the revolution of TN ( = 8s)
about AB ; its area is of the form 27rySs and the intensity of
illumination is OC, say. The length y is the projection on the
horizontal axis through O of either the line OT, the line ON, or the
line midway between these (for these differ in length but slightly
if 8s is taken as very small), i. e., y = OT cos 6
or y .= Rcos0.
264 MATHEMATICS FOR ENGINEERS
Hence, for this zone, the illumination
= candle power x area
= OCX27rRcos0Ss
= a&X27rRxfl«'ifor-r- = cosflf •
v OS )
Hence the total illumination
= 27rR X area under the curve FL&D
and thus
Total illumination 2?rR X area under the curve FL&D
M = 47rR2 47rR2
_ area under the curve FL&D
~^zKT
= mean height of the curve FL&D
since 2R is the base of the curve.
Dr. Fleming's Graphic Method for the Determination of
R.M.S. Values. — The determination of R.M.S. values is of some
importance to electrical engineers,
and the subject, previously dis-
cussed in Chapter VII, is here
treated from a different aspect.
Instead of squaring the given
values of the current and then
extracting the square root of the
mean of these squares, we may,
by a simple graphic construction,
obtain the mean of the squares very
readily.
Let the values of an alternating
current at various times be as in the table : —
time t . .
o
•001
•002
•003
•004
•005
•006
•007
•008
Current C
5
8
12
7
0
-6
-8-3
-3
5
then, to find the R.M.S. value of C we proceed as follows : —
Treat the given values as polar co-ordinates, taking t for the
angles and C for the rays. Select some convenient scale for t, say
20° — -ooi sec., and a scale for C, say i"=4 units, these being the
scales chosen for the original drawing of which Fig. 99 is a copy to
POLAR CO-ORDINATES
265
about one-half scale; and set out a polar diagram as indicated,
making OA = +5, Oa = —6, etc. Join the extremities of the rays,
so obtaining, with the first and last rays, the closed figure ABCDaE.
Measure the area of this figure by means of a planimeter — in this
case the area was found to be 4-23 sq. ins.
Now the area of the figure = %frzd& = %fC2dt, or
so that if we divide twice the area by the range in t, the mean
value of the squares is determined.
In this case the range of t = 160° = 2-79 radians, and also
J"C2dt = 2 Xi6x 4-23 = 135-5, f°r I"=4 units, and thus i sq. in.
= 16 sq. units.
Then
M.S. =
279
= 48-65,
and hence R.M.S. =
= 6-98.
The rule for the area of a figure, viz., %J"r2dO, may be usefully
employed to find the height of the centroid of an area above a
certain base.
Example 5. — Find the height above the base OX of the centroid of
the irregular area OABX [(a) Fig. 100].
fir
10
a
r
B
8 12 16 EO
(O
FIG. 100. — Centroid of Irregular Area by Polar Diagram.
266 MATHEMATICS FOR ENGINEERS
To do this, first divide the base into a number of equal divisions
and erect mid-ordinates in the usual way. Measure these mid-ordinates
and set off lengths to represent them as radii vectors from A in (b)
Fig. 100, the angles at which the rays are drawn representing the
values of x, i. e., the lengths of the divisions of the base. Join the ends
of these rays and measure the area of the polar diagram thus obtained ;
divide this area by the area of the original figure and the result is the
height y required.
For the area of the polar diagram = %fr2 dd = \fyz dx, since rays
represent values of y and the angles represent values of x. Also the
area of the original figure =fy dx, so that
area of polar diagram _ \fyz dx _ _
area of original figure fydx ~
For the particular case illustrated (the scales referring to the original
drawing) : —
For (a) Fig. 100 i" = 5 units vertically
i* = 4 units horizontally
so that i sq. in. represents 20 units of area. The area was found by
the planimeter to be 16-82 sq. ins., so that the actual area is 336-4
sq. units, i. e., fy dx = 336-4.
For (b) Fig. 100 i" = 5 units radially
and each angular interval = 20°, so that the total range = 180° or 3-14
radians. Hence 3-14 radians represent 20 units, the length of the base
in (a) Fig. 100,
or i radian = — 6-36 units.
3-14
Now the area is of the nature r2X0, i. e., (length) 2x angle, hence
i sq. in. of area = 52X 6-36 or 159 units. Area of the polar figure
(found by the use of the planimeter)
= 18-34 scl- ms- = J8'34 X 159 units = 2920.
Hence y = ^|£. = 8-68,
336-4
i. e., the centroid horizontal is found.
Theory of the Amsler Planimeter. — The principle upon
which the planimeter is based may be explained quite simply, in
the following way.
In Fig. 101 let PP" be a portion of the outline of the figure
whose area is to be measured, and let the fixed centre of the
instrument be at O. Then in the movement of the tracing point
P from P to P" along the curve, the tracing arm changes from the
position AP to A'P". This movement may be regarded as made
up of two distinct parts : firstly, a sliding or translational move-
ment from AP to A'P', and next, a rotation round A' as centre,
POLAR CO-ORDINATES
267
from A'P' to A'P". In the former of these movements the record-
ing wheel moves from W to W, but part of this movement only,
viz., that perpendicular to the axis of the wheel, is actually
recorded, so that the wheel records the distance p.
The area swept out by the tracing arm AP during the small
change from P to P" = APP'P"A' = APP'A'+P'P'A'
Hence for the whole area,
area swept out = 2APx/>+2£(AP)2S<9
Now the net angular movement is zero, so that 280 = o .
Hence area swept out = AP2/>,
or if / = length of the tracing arm,
area = / X travel of wheel
and hence the reading of the wheel
_ area of figure
Thus the length of the tracing arm determines the scale to
which the area is measured. Hence by suitable adjustment of this
length of arm the area of a figure may be
read in sq. ins. or sq. cms. as may be neces-
sary. If the average height of the figure is
required, the length of the tracing arm must
be made exactly equal to the length of the
figure. This is done by using the points LL
(Part I, Fig. 301), and not troubling about
the adjustment at A. The difference between
the first and last readings gives, when multi-
plied or divided by a constant, the actual
mean height of the figure. If the ordinary
Amsler is used, then the mean height in
inches is obtained by dividing the difference
between the readings by 400 ; thus if the first
reading was 7243 and the last 7967, the mean
height would be the difference, viz., 724,
divided by 400, i.e., 1-81 ins.
The area of the figure = average height X
length, but the area of the figure = length of
tracing arm X wheel reading, hence if the length of the tracing
arm = the length of the diagram, the wheel reading must be the
average height of the diagram.
FIG. 101. — Theory of
Amsler Planimeter.
268
MATHEMATICS FOR ENGINEERS
[It should be noticed that the area recorded by the instrument
is really the difference between the areas swept out by, the ends
A and P of the arm AP, but as A moves along an arc of circle,
coming back finally to its original position, no area is swept out.]
Exercises 21. — On Polar Co-ordinates.
1. Plot a- polar curve of crank effort for the following case, the
connecting rod being infinitely long.
e°
o
15
3°
45
5'i
60
75
90
7'3
105
1 20
i35
. 150
165
1 80
Crank Effort (Ibs.) . .
o
2-4
3-9
6-4
7-1
7
6-1
4'9
3'3
»*5
0
2. As for Question i, but taking the connecting rod = 5 cranks.
e°
0
15
3°
45
60
75
90
i°5
1 20
5-8
'35
'5°
165
172-5
i So
C.E.flbs.) ....
0
2<45
4'4
6-1
TO
7'4
7'4
6-6
4'4
3'°
i'S
•6
0
3. An A.C. is given by C = 7-4 sin ^oirt. Draw the polar curve to
represent the variation in C and hence find its R.M.S. value.
4. What is the polar equation of a circle, the extremity of a
diameter being taken as the centre from which the various rays are to
be measured ?
Of what curve (to Cartesian ordinates, i.e., rectangular axes) is the
circle the polar curve ?
5. Plot the polar diagram for the arc lamp, from the table.
Angle (degrees) . . .
o
TO
2O
3°
40
50
60
70
80
90°
(vertical)
Candle Power ....
800
T2OO
l6<X)
2OOO
2200
2200
2300
2500
2300
1800
6. Plot the Rousseau diagram for the arc lamp in Question 5 and
from it calculate the M.S.C.P. of the lamp.
7. An A.C. has the following values at equal intervals of time : 3, 4,
4'5> 5'5> 8, 10, 6, o, —3, —4, — 4-5, —5-5, —8, — 10, — 6, o. Find by
Dr. Fleming's method (cf. p. 264) the R.M.S. value of this current.
8. Eiffel's experiments on the position of the centre of pressure for
a flat plane moved through air at various inclinations gave the follow-
ing results : —
Inclination to horizontal ....
0
5
10
15
3°
45
60
75
90
T>p
Ratio (see Fig. 102) .....
•263
Plot a polar diagram to represent the variation of this ratio.
POLAR CO-ORDINATES
269
9. Draw the polar curve to represent the illuminating power of a
U.S. standard searchlight from the following figures : —
Angle (degrees)
o
(vertical)
10
20
3°
40
5°
60
70
80
90
Candle Power .
3000
10000
20500
33000
41500
4*5°o
43000
43000
30000
24000
10
(above
horizontal)
20
30
40
50
60
70
80
90
9000
6000
5000
5000
2OOO
3000
1500
1500 I50O
C 15 centre of pressure
FIG. 102.
CHAPTER IX
SIMPLE DIFFERENTIAL EQUATIONS
Differential Equations. — An equation containing one or more
derived functions is called a " differential equation."
Thus a very simple form of differential equation is
dy =
dx ?
dv
— 5 = o
x
, dy
and +2x-=°
and 4 -;- + 7/ — 5
dx2 dx
are more complex forms.
Differential equations are classified according to " order " or
" degree " ; the order being fixed by that of the highest differential
£/V
coefficient occurring in it. Thus ~ is a differential coefficient of
dsy
the first order. -—• is of the third order, and so on.
dx3
dv
Hence 4-2 -\-y-jr = 5 '34 is an equation of the first order
w%
d*y
and 8-^j+.y = 7-I6 is an equation of the fourth order.
wsv
The "degree" of an equation is fixed by that of the highest
derivative occurring when the equation is free from radicals and
fractions.
d2y
Thus T-TJ = c is of the second order and of the first degree
whilst 4(3-^) +(-r"o) = 7 is °f the second order and of the
2 2
second degree.
Much has been written concerning the solution of the many
types of differential equations, but it is only possible here to treat
270
SIMPLE DIFFERENTIAL EQUATIONS 271
the forms that are likely to arise in the derivation of the proofs of
engineering formulae ; the plan being to discuss the solution accord-
ing to the types of equation.
Type : — ~ given as a function of x. — With the solution
of such simple forms we have already become familiar, for the
equations connecting the bending moment at various sections of a
beam with the distances of those sections from some fixed point are
of this character.
Thus taking the case of a simply supported beam carrying a
load W at the centre
dfy = \V
dx* ~ El
dy Wx
whence -£- = ^F +C
dx El
which is of the type under consideration.
Evidently this equation can be solved by integration through-
out, attention being paid to the constants which are necessarily
introduced. Expressing in algebraical symbols,
!=/<*> ;
then by integrating throughout with regard to x
or y = \f(x}dx-\-C.
*v
Example i. — If -~ = ^xz+rjx— 2 and y = 5 when x=i, find an
a%
expression for y in terms of x.
This equation is of the type with which we are now dealing, since
— 2 =/(*)
Integrating y = —r - + * -- 2X+C.
The value of C must now be found : thus y = 5 when x = i
so that 5 = A_(_i_ 2+C
or C = 2-iy.
Hence y= i-33#3+3-5#2— 2^+2-17.
272 MATHEMATICS FOR ENGINEERS
Type : — f- given as a function of y. i. e., -^ — f(v).
ax dx
This type of equation differs somewhat from the preceding in
that a certain amount of transposition of terms has to be effected
before the integration can be performed.
dy
The equation may be written 7— = dx
the transposition being spoken of as " separating the variables,"
and thence by integration I -,-,--. — \dx-\-C = x-\-C.
jj(y) J
dy
Example 2. — If ~ = y3, find an expression for y.
ax
Separating the variables — ^ = dx.
Integrating / -^ = / dx-\-C
or —\y-* = x+C
whence x-\ — - + C = o.
y _
The two foUowing examples are really particular cases of the
type discussed generally on p. 275, but they may also be included
here as illustrations of this method of solution.
dy
Example 3. — Solve the equation -j- + ay = b.
(tX
dy ,
Here -g = b— ay
dy
J = dx
so that ^- = tdx+C
— J
T
b— ay
^-
b—ay
i. e., loge (b—ay) = x+C
or loge (b—ay) — — (ax-}-aC)
whence e~ax-aC = (b-ay).
Now let A = e~aC: then e-«*-«c = e~ax x e~aC = Ae~<
and Ae~aa:— & = —ay
b A _
or "V = £ •
^ a a
SIMPLE DIFFERENTIAL EQUATIONS 273
dy
Example 4. — Solve the equation 4-^- = u-f-7y.
Separating the variables
i.e.,
or loge (u+jy) = lx+
whence e*x+*C = (11+77)
1C
and if A be written in place of e
11+77 = Ae*z
A \x ii
or y = — e ---
* 1 7
Example 5. — The difference in the tensions at the ends of a belt
subtending an angle of dd at the centre of a pulley = dT = Tpdd, where
/* is 'the coefficient of friction between the belt and pulley. If the
greatest and least tensions on the belt are T and t respectively, whilst
T
the lap is 6, find an expression for the ratio — .
The equation dT = Tfjdd is of the type dealt with in this section ; to
solve it we must separate the variables, thus : —
Integrate both sides of the equation, applying the limits t and T to
T and o and 6 for the angle.
/"T dT re
Then — =M/ 06.
J t i Jo
/ \T / \fl
Mr <•(').•
T
But loge- = loge T - loge /•
T
Hence *O£' x ^
or
A word further might be added about Example 3, or a modifi-
cation of it.
T 2
Let -f- = ay.
dx *
T
274 MATHEMATICS FOR ENGINEERS
If a = i, then -~ =y, i. e., the rate of change of y with regard to
ctoc
x, for any value of x, is equal to the value of y for that particular
value of x. Now we have seen (Part I, p. 353) that this is the
case only when y = ex.
If a has some value other than i, y must still be some power of
e, for the rate of change of y is proportional to y ; actually, if
-V = 6°*, -¥- = aeax = ay, so that y = e0* would be one solution of
dx
the equation -¥. = ay, but to make more general we should write
d%
the solution in the form y = eax-\-C or y = Aeax, whichever form is
the more convenient. Whenever, therefore, one meets with a
differential equation expressing the Compound Interest law (*'. e.,
when the rate of change is proportional to the variable quantity)
one can write down the solution according to the method here
indicated.
Example 6. — Find the equation to the curve whose sub-normal is
constant and equal to •za.
dv
The sub-normal =)>-/-• (See p. 43.)
dx
dv
Thus y~- = ia,
Jdx
or, separating the variables, fydy =fzadx.
V2
Hence — = 2ax+C (Integrating)
or -y2
This is the equation of a parabola ; if y = o when x = o, then K = o
and y2 = ^ax, i. e., the vertex is at the origin.
Example 7. — Find an expression giving the relation between the
height above the ground and the atmospheric pressure ; assuming that
the average temperature decrease is about 3-5° F. per 1000 feet rise, and
the ground temperature is 50° F.
Let T be the absolute temperature at a height h,
then, from hypothesis
r = 460+ 50 — =-£• h
1000
= 510 — 0035/1 (i)
Now we know that pv = CT (2)
and also that if a small rise 8h be considered, the diminution in the
SIMPLE DIFFERENTIAL EQUATIONS 275
pressure, viz., 8p, is due to a layer of air 8h feet high and i sq. ft. in
section, and thus
From equations (i) and (2)
pv = €(510— -0035^)
and substituting for v its value from equation (3)
or, in the limit
dp i
pdh ~ €(510 — 0035/1)
Separating the variables
dp _ dh
p "€(510 — 0035/1)
Integrating, the limits to p being p0 and p, and those to h being
o and h,
(-logpY = £x — -[log (510— 0035 A) -log 5IO|
\ /p0 C —-00351- J
whence log p— log p0 = ~ [log (510 — 0035/1) -6-234J
or log p = log Po+^ [_log (5IO-'°°35A)— 6-234J
which may be further simplified by substituting the values for p0 and C.
General Linear Equations of the First Order, i. e.,
equations of the type
where a and b may be either constants or functions of x.
The solution of this equation may be written as
y =
The proof of this rule depends upon the rule used for differen-
d(uv) du , dv
tiatmg a product, viz., ~-± =v,--\- u~r- ; the reasoning being as
dx dx &x
follows : —
Let us first consider the simplest case in which this type of
equation occurs, viz., the case of the solution of the equation
dy
where a is a constant.
276 MATHEMATICS FOR ENGINEERS
Multiplying through by £«* the equation becomes
dy ,
e^~--\- aye** = o,
dx
dy dv
which can be written as v^- + v— = o (where v = e°* an(j thus
dx dx
*--•»•
But v-r 4- V-T- = ~r(yv)» s° that -4-- (vv) = o ; hence w must be
dx dx dx^ Wfw
a constant, since the result of its differentiation is to be zero.
Accordingly yv = C,
or y = Ctr1 = Ce~ax.
Extending to the case in which 6 is not zero, whilst a remains a
constant, i. e., the equation is
dy
»+«"*
we find that after multiplying through by &** the result arrived at is
d
Integrating both sides with regard to x,
ye** = fbe^dx+C
or y=e-flJC{fbeaxdx-\-C}.
This may be evaluated if the. product of b and e** can be
integrated.
For the general case, that in which a and b are functions of x,
the multiplier or integrating factor is e^adx, for after multiplication
by this the equation reads
dv
efadx y J^.aejadxy = fefadx
and this may be written
whence by integration we find that
yefadx =
or y =
SIMPLE DIFFERENTIAL EQUATIONS 277
dv
Example 8. — Solve the equation j^+izy = e*x.
The equation may be written
dy 12 i
j-H — y = -e**
** 7* 7
so that in comparison with the standard form
a = — and b = -e**.
7 7
Hence y = «-/JT***{ fie**. es'
-7 47
—
47
Example g. — If -j-— y = 2x+i, find an expression ior y.
dv
so that a= — i, 6 =
Hence y = e+f<'*{f(2x+i)e-J<**dx+C}
* The value of the integral f(2x+i)e~*dx is found by integrating
by parts.
Thus, let (2x+i)e~x = udv where dv = e~x, i. e., v = — e~x and
. du
U=2X+I, t. e.,-j- = 2,
then fudv = uv—fvdu
= [(2x+i)x(-e-*)]-f-e-*2dx
/FT
Example 10. — Solve for T the equation -3 — |-PT = P(t—cx) (referring
to the transmission of heat through cylindrical tubes) ; P, t and c being
constants.
278 MATHEMATICS FOR ENGINEERS
The equation -^ — [-PT — P(t—cx) is of the type -/--{-ay = b, where
(A/X dX
a = P and b = P(t—cx).
Hence the solution may be written
T = e-SPd*(fP(t-cx)efpdxdx+K},
the integrating factor being e^7dx, i. e., e^x.
Hence T = e -Vx{fP(t-cx)ePx dx+K} ;
and to express this in a simple form the integral fP(t— cx)ePx dx must
first be evaluated.
Let fP(t—cx}epxdx=fudv where u=P(t—cx) and dv = ePx dx so
that v — ^f?x ', and also du = —Pc dx,
then fP(t—cx)epxdx =fudv = uv—fvdu
= P(t-cx) x *Pa:+i x e**.Pcdx
r rp -,
T = e-px\(t-cx)e**+^- +M+K I
since e~fxxePx — e° = i and L — M+K.
Example 1 1 . — When finding the currents x and y in the two coils of
a wattmeter we arrive at the following differential equation :
dy , R,+R2
where Rx and Lj are respectively the resistance and inductance of the
one coil and R2 and L2 are the resistance and inductance of the other
coil ; I being the amplitude of the main current.
Solve this equation for y.
/1<\)
Comparing with the standard form of equation, viz., -^--{-ay — b, we
ii . -"-^-l I JLVrt . . J-/l A' •*- , t i -LV1 * • , i
see that a — x * and o = ~_, cos ^>^+T _^T sin />^-
Hence
/Rl~f~R2j. /• r / T J^T TI> T
T , T — at I // -Lipl , . . -tC,!
l/l + jjj -( I I ii r.O«V5<-l i—
\J VLi+L,
/2
•- /• T J,T
J / /y_- cospt.e
IJ Lf-f-1/
(Ri+R2X
r Tj r i2 -,
+ / * sinpt . e Li+L2 dt+C \
J L,i-\-L,2
=e~Atl fB cospt . eAtdt+ /D sin^ . eAtdt+C\.
SIMPLE DIFFERENTIAL EQUATIONS 279
Now, as proved on p. 157,
I B0A« cos pt dt = * (p sin pt-\- A cos pt) -f Cx
and j ~De A* sin pt dt = A>1 *, (A sin £* — £ cos £*) + C2
Hence
- At x e At ( _
sin ^i+ AB cos pt+DA sin pt—Dp cos
A2. .a
= * ]
R,+R2 lL.pl R.I
Exact Differential Equations. — An exact differential equation
is one that is formed by equating an exact differential to zero ; thus
Pdx-}-Qdy = o is the type, Pdx-{-Qdy being an exact differential.
The term exact differential must first be explained.
Pdx -\-Qdy is said to be an exact differential if — = ^S, the
3y dx
derivatives being partial, or, to use the more familiar notation,
dy / \dx.
To solve such an equation proceed as follows : If the equation is
exact, integrate Pdx as though y were constant, integrate the terms
in Qdy that do not contain x, and put the sum of the results equal
to a constant.
[For, let Pdx+Qdy = du.
Now, du is the total differential, ( -=- }dx and ( -j- }dy
\dxJ \dyJ '
being the partial differentials (see p. 82) ;
(du\ (du\
i. e., du = ( -=- MX+ I -j- )dy.
\dx/ \dy/ "
TU -t ^ (du\j i (du\j
Then if du = o, ( — ]dx4-\ -3- jay = o,
\dx' \ay/ '
and this is exactly the same as the original equation
., (du\ idu\ „
if ( -j- } = P and ( -=- } = Q,
\dx/ \dy/
^y/ \iy/\dx/ \dy.dx
\ (d\(dii\ (dQ
I -=— )\ -.— I
dx.dyJ \dx/ \dy / \dx
28o MATHEMATICS FOR ENGINEERS
Our equation thus reduces to du = o, or, by integrating, to u — C,
but u= I Pd% (y being constant) + I Qdy (x being constant), and
hence we have the rule as given.]
Example 12. — Solve the equation
(x2— 4%y— 2y2) dx+(y2—4xy—2x2) dy = o.
Here P = xz-^xy-2y2 (^P) = -^x-^y
\dy 1
Q = y*-4xy-2x*
and thus the equation is exact.
I P dx (as though y were a constant)
/X^ 4X2V
(x2—4xy—2yz) dx = 2--i— 2y*#.
I Qdy (as though x were a constant)
= {(y*—4xy—2xz) dy = ^*— ^ — 2Ar2y ;
but of this only ^3 must be taken, since the other terms have been
obtained by the integration of terms containing x.
Hence \x*— 2x2y— zxyz+$ys = C
or x3— 6x*y—6xy*+y* = K.
Example 13. — Solve the equation v du— u dv = o.
If this equation is multiplied through by - 2 we have a form on the
left-hand side with which we are familiar, viz.,
v du—u dv _
v2 ~~°'
for the left-hand side is d ( - V
It ••
Then by integrating, - = C
or u = Cv.
I This equation might have been regarded as one made exact
through multiplication by the integrating factor — 2.
SIMPLE DIFFERENTIAL EQUATIONS 281
Equations Homogeneous (/. e.. of the same power
throughout) in x and y.
Rule. — Make the substitution y — vx and separate the variables.
Example 14. — Solve the equation (xz+yz) dx = 2xy dy.
Let y = vx,
dy dx , dv . dv
then -f- = V-J-+X-T- = v-\-x-r ..... -. . (i)
dx dx dx dx
Now (xz+yz) dx = 2xy dy
dy _ xz+yz _ x*+xzvz _ i+vz
~~" -- - -
SO , ~~" «
dx 2xy 2xzv
Substituting for -- from (i)
or
dx -2V
dv ijrvz—2v2 i
3- — - ---
dx -2V 2V
Separating the variables, and integrating,
fzv dv _ i~dx
J i—vz~J x
i. e., — log (i— vz) — log#+log C rthe substitution being
u = i— v2
or log#(i— vz) = — logC = log K I du=—2vdv
i.e., x(i—vz) = K
or
and xz—y* =
Linear Equations of the Second Order.
Let v - «**, then - Ae^ and -
«X ^2
so that XV^+aAe^+fce** = o
or X2+«X+6 = o for jy = o would be a special case.
There are three possible solutions to this quadratic.
282 MATHEMATICS FOR ENGINEERS
The general solution is : X = = —
^ i v az — 46
and let X, =
We shall now discuss the three cases.
Case (i). — If #2>4&, then Xj and X2 are real quantities and
unequal.
d V tt"V
Now if v = A, e^x. -r4 + a-r- + by will equal o, as would be the
1 dxz dx
case also if y = A2e^, so that to complete the solution the two
must be included (for the equation is true if either or both are
included).
Thus y = Ajtf*!* -\- A%e^2X
the constants Aj and A2 being fixed by the conditions.
Case (2).— If A2 = 46, then X1 = X2.
According to the preceding case we might suppose that the
solution was y = Ae^.
This, however, is not the complete solution,
which is y — (A+B*)^.
Case (3). — If az<4b. This means that Va2— 46 is the square
root of a negative quantity, i. e., it is an imaginary.
Now, a2— 4& = —1(46— a2), (46— a2} being positive ;
hence Va2— 46 = V^i Vtf—a2
—a+j\/4b—az
and
— a— A—
Use might be made of the solution to C<zs0 (i), adopting these
values of Xa and X2, but this does not give the most convenient
form in which to write the solution.
Let c=V4&— az then Xj =
x
— rt — ;c
and X, = i-
SIMPLE DIFFERENTIAL EQUATIONS 283
Then y = A^i^-f- A2e*& from Case (i)
(-a+jc)x (-a-jc)x
Developing one of these only, viz., the first, and neglecting
for the time being,
-ax+jcx _ax jcx
e 2 =e~ 2"x«~2".
Now ei* = cos x -\-j sin x (see p. no),
and by writing — for x
& cx, . cx
e2 = cos — \-ism —
2 2
Hence
"" ' cx , . cx\ , -¥/ cx . . cx\
cos — +7 sm — + A.2e 2 / cos j sin — )
< 2 2 / \ 2 2 /
where
A = V(A1+Aa)a+y*(A1— A2)2 = 2 VA^AZ . (see Part I, p. 277)
and
Taking as the standard equation
and grouping our results, we have the following
(i) If a2>4&: the solution is
y == Al6
(2) If a2 = 46 : the solution is
_ax
y= (A+Bx)e ~*.
(3) If az<4b: the solution is
284 MATHEMATICS FOR ENGINEERS
The last of these forms occurs so frequently that very careful
consideration should be given to it, and to the equation of which
it gives the solution.
Example 15. — Solve the equation
dzy , dy
^- — 2V = O.
dx f
This can be written (after dividing through by 5)
dzy , dy
-3-4+2-4-^ — 4V = o,
dx2 dx
so that a — 2-4 and b — — -4 (in comparison with the standard form).
-2-4+v/5776+l:6,|. -2'4-v/6-76+l-6.
y = AlS
It is really easier to work a question of this kind from first principles
rather than to try to remember the rule in the form given ; thus the
values of X will be the roots of the equation
5\2+ I2X— 2=0.
Then, calling these roots Xj and X2 respectively
If the values of At and A2 were required, two values of y with the
corresponding values of x would be necessary.
Example 16. — A body is moving away from a fixed point in such a
way that its acceleration is directed towards that point, and is given in
magnitude by 64 times the distance of the body from that point. Find
the equation of the motion and state of what kind the motion is.
The motion is Simple Harmonic. (See p. 60.)
If s = displacement at time t from the start
= acceleration and = — 645
(the reason for the minus sign being obvious).
dzs dzy dy
Thus ^+645=0, which is of the type -^-}-a~-\-b=o, where a=o
and 6=64.
If s = e*
X2 + 64 = o
so that X = ± V— 64= ± 8j.
SIMPLE DIFFERENTIAL EQUATIONS 285
Hence the solution (according to Case (3)) is
_??
y — Ae 2 sinf -- \-p\
where a = o and c — V^b—a* — 16
so that s = A sin (8t+p) .
The general equation of S.H.M. is
s = A sin (o>t-}-p)
&> being the angular velocity, so that the angular velocity in this case
is 8 and the amplitude is A.
Example 17. — Solve the equation
Let y = e**
then X2+8X+i6 = o
i.e., (X+4)2 = o
i. e., the roots are equal.
Hence y = (A+~Bx)e>&, (Cf. Case (2), p. 282)
where X = — 4.
hence y = (A-\-~Bx)e~*x.
Example 18. — Solve the equation
dzv dy
This differs from the preceding examples in the substitution of a
constant in place of o.
The equation can be written
dzy , dy .
5^+7i+io(y--5)=°.
Let (y—~5) = e*x
then (Q. = \ete and ^
dx dxz
and X2-|-7X-j-io = o
whence X = — 5 or — 2
then y- -5 = A^ - fa+ Aze
or = Ag-
//y /-y
In other words, the solution is that of -=^j + 7 -— + ioy = o plus
tt# i**V
5 . the constant dy j ^
— , t. e., - ^^ -- T-. This is correct because, if v = -5, -^ and -v4
10 coefficient of y dx dx2
each equals o, and thus one solution is y — -5. The complete solution
is the sum of the two solutions.
286 MATHEMATICS FOR ENGINEERS
The Operator D. — The differential coefficient of y with
respect to x may be expressed in a variety of forms : thus either
—-, -j , f'(x) or Dy might be used to denote the process of
differentiation. The last of these forms, which must only be used
when there is no ambiguity about the independent variable, proves
to be of great advantage when concerned with the solution of
certain types of differential equations. It is found that the symbol
D has many important algebraic properties, which lend them-
selves to the employment of D as an " operator."
The first derivative of y with regard to x = Dy, and the second
d*y
derivative of y with regard to x — -j^, which is written as D2y ; D2
indicating that the operation represented by D must be performed
twice. This is in accordance with the ordinary rules of indices, so
the fact suggests itself that the operator D may be dealt with
according to algebraic rules. Thus D3 must equal D.D.D (this
implying not multiplication, but the performance of the operation
three times) ; for
..
dx* dx\dx2/ dx dx dx
Our rule then holds, at any rate, so long as the index is positive,
or the operation is direct ; and for complete establishment we must
test for the case when the index is negative.
If Dy = dj? D = f:let^ = m, i. *., Dy = m.
Then by integration y = I mdx,
but if Dy = m and the rules of algebra can be applied to D
m i
y must = ^r or =- . m.
Hence -^m— Imdx
)W~ \i
or =- indicates the operation of integration.
Again, if the rules of indices are to hold,
D.D"1^ must = D°y or y,
hence D-1 must represent the process of integration; since if we
SIMPLE DIFFERENTIAL EQUATIONS 287
differentiate a function we must integrate the result to arrive at
the original function once again.
Hence D'1 = ^.
Having satisfied ourselves that the ordinary rules of indices may
be applied to D, we may now prove that the rules of factorisation
apply also.
Taking the expression D2 — 120+32, we can easily show that it
can be written in the factor form (D— 4)(D— 8) :
for let y = jx2—5x, then Dy = 14*— 5 and D2jy = 14.
Also (D2— 120+32)? = D^y— i2Vy+32y
= 14— i68#-f 6o+224#2— 160*
and
— 5— 56*2+40*)— 4(14*— 5—
= 14 — II2#+40 — 56#+20+224#2 — I6OX
so that
(D2-I2D+32) = (D-4)(D-8).
These properties make D of great usefulness in the solution of
certain types of differential equations : e. g.,
Suppose 5^+7^ + ioy = M ..... . (i)
then this equation may be re-written as
M5D2+7D+io)=M
M
y =
5D2+7D+io
and the solution of equation (i) may be found by this artifice.
Many differential equations occurring in electrical theory may be
solved in a very simple manner by the treatment of D as a " quasi-
algebraic" quantity: before proceeding to these, however, we
must enunciate the following theorems.
288 MATHEMATICS FOR ENGINEERS
Useful Theorems, involving the Operator D : —
(1) (p+qD) operating on the function a sin (bt+c)
gives the result aVp2+b2q2 sin f^+c+tan"1^ )
\ p '
(2) r—p. a sin (bt+c) = ——_ - sin (bt+c— tan-1-).
p+qV ^*+b2* P'
Proof of (i).
(p+qD)a sin (bt+c)
= ap sin (bt+c)+aqb cos (bt+c}
= aVp*+q*bz sin (bt+c+ia.^^ (See Part I, p. 277.)
\ p/
Proof of (2).
sn
J j> sin (U+c)—bq cos
J
"1
sin
a sin ( 6^+c— tan-1-
V
As a test of the correctness of the above rules the combination
of the two operations should give the original function.
Thus
p-\-qu
sn
= a sn
A third theorem might thus be added.
D sin (bt+c) = 6 cos
D2 sin (&^+c) = — Z>2 sin
or D2 = -62
hence pz—q2Dz = p2+qzbz.
SIMPLE DIFFERENTIAL EQUATIONS 289
Application of these Rules to the Solution of Differential
Equations.
d^y dy
Example 19. — Solve the equation -5-^+7^4-12^ = eSx.
This equation might be written
S0that ^
The solution of this equation gives the particular integral, whilst
the complementary function, as it is termed, will be obtained by the
solution of the equation
dx* dx
[The solution of this equation we know from the previous work to be
Now T>e5x = 5eSx, DV5* = 250^, i. e.t D = 5 and D2 =
Hence the particular integral is
~ 25+35+12
^g5*
~7*'
Hence the general solution is
To test this by differentiation of the result : —
2e-*;t+— e&x
= e*x.
dzs ds
Example 20. — Solve the equation ,- +4^7+45 = 5 sin 7/. (This type
at dt
of equation occurs frequently in electrical problems and in problems on
forced vibrations of a system.)
U
290 MATHEMATICS FOR ENGINEERS
/72 c /7c
The solution of --2+4-^7+45 = 0
(It Cli
is s = (A+Bt)e~zt (See p. 283.)
To find the particular integral : —
(D2+4D + 4) 5 = 5 sin jt.
_5sin jt
-DM-4D+4
D sin jt = j cos jt and D2 sin jt = — 49 sin jt.
(Note that D2 = —49, but D does not = j.)
We must thus eliminate D from the denominator: to do this,
multiply both numerator and denominator by D2+4— 40.
Then _5(P2+4-4P)sin7/
(D2+4)2-i6D2
_ 5(— 49 sin 7^+4 sin jt—28 cos jt)
(-49+4)2-(i6x-49)
_ —5(45 sin 7^+28 cos jt)
2809
_ 28
45
2809
Hence the complete solution is
53 A' ' in~14-5>
53 ~~V ' "dn 45A
The particular integral might have been found in a more direct
fashion by the use of Theorem 2, p. 288.
For
sin7* i
(D+2)2-(D+2) •
(7\ (4) = 2
jt— tan"1-), _,,
_ 2j from Theorem 2, J ? =
sinf jt— tan-1-J
p. 288 I c = o
U = 7
I I / 7 7
= —7= X — 7= sin 1 7i— tan"1 - — tan-1 -
V53 V53 V 2 2
= — sin ( jt— 2 tan"1 - Y
53 V 2}
SIMPLE DIFFERENTIAL EQUATIONS 291
The two results do not appear at first sight to be the same, but the
can be reconciled in the following manner : —
tan"1 — = tan"1 -6223 = 31° 54'
tan-1 ? =74° 3'
Thus — sin (7/4- tan-1 — ) = — sin (7/4-31° 54')
\ 45'
also sin (7*— 2tan~1-J = sin (jt— 2x74° 3') = sin (jt— 148° 6')
= — sin (1804-7^— 148° 6')
= - sin (7/4-31° 54')
= — sin ( 7/4- tan"1 — ).
v 45/
IT Tl, -.-' T^T&y , T- , B/ 1TX
Example 21. — The equation EI-r^4-Fy4--^- cos-=- = o occurs in
dx o I
Mechanics, y being the deflection of a rod of length /, and F being the
end load.
Solve this equation.
,,T dzy , F B/ TT*
We may rewrite the equation as -r^+^fV =— B^F cos -; • • I1)
dxz EF 8EI /
The solution of +-^y = oisy = Asui/\ j^x+p ... (2)
Reverting to form (i)
B/ nx
B/
or 8EIC<S
B/ irX
8-cos
F
B/
Hence the complete solution is y = A sin ( *J ~pX-{-p j4-
8 COST
292 MATHEMATICS FOR ENGINEERS
Example 22. — A pin-jointed column, initially bent to a curve of
cosines, has a vertical load W applied to it. Find an expression for
the deflection at any point.
Given that the equation of the initial bent form is
irX
y being the deflection at distance x from the centre of the column,
which is of length /.
Also
this equation being obtained from a consideration of the bending
moment at distance x from the centre.
,
and thus +--A cos
W AW fnx
iLiy= -ETCOS
firX\\
(y J) = o
fnx\
(T>
Now, as shown on p. 283, the solution of the equation
d* W
W
To find the particular integral, viz., the solution of the equation
d* . W AW
write the equation as
AW
AW /irX\
— -^-= COS ( — )
so that « = - \l >
AW l-nx
—
El I*
Combining the two results
(r).
SIMPLE DIFFERENTIAL EQUATIONS 293
The following example combines the methods of solution
employed in Examples 17, 18, 19 and 20.
Example 23. — Solve the equation
d2s ds
= «-«+ sin 607T/+5.
(a) The solution of
-5^— 12-^ + 205 = 0 is s = A.1elot+Azezt.
(b) The particular integral for
d?s ds
d7*-12di
may be thus found : —
D2— I2D+20 D2e~5t = 2$e-
i
25+60+20
(c) To find the particular integral for
d*s ds
-jr.— i2^-+2os = sinooiri.
dtz dt
D2— I2D + 20
I ^ sin6o7r<
D— 10 D— 2~
^ • ~/'=\ ==,; sin [6o7rf— tan"1 (—3071-)] \q = i J-
D-io V4+36007T2 l& = 6o^J
i i
/- -.- , -^ ' -/— ^~z — a sin [607T/— tan"1 (—30^)— tan"1 (— 6rr)l
v4+3ooon-a Vioo+3ooo7ri
sin [6orr<— tan^1 (—3077)— tan"1 (—677)]
2oV(l+9007T2) (I + 367T2)
(d) The particular integral for
dzs ds
is
20 4
Hence the complete solution is the sum of those in (a), (b),-(c)
and (d), viz.,
s = A e"'+A g^|g~5
1 2
105 _ 2oV(i+90on2) (I+367T*)
294 MATHEMATICS FOR ENGINEERS
Equations of the Second Degree. — The treatment of these
equations is very similar, up to a certain point, to that employed
in the solution of ordinary quadratic equations ; particularly the
solution by factorisation.
Example 24. — Solve the equation
^Y- 8^-33=0
dx/ dx
Let Y= then Y2-8Y~33=o
dx
or (Y-ii)(Y+3) = o.
Thus Y = 1 1 or Y = — 3
dy dy
i.e., / = ii or ^-= — 3,
dx dx
whence y = i ix + Ct or y = — $x + C2
or y — i ix — Cx = o y + 3, — C2 = o,
and the complete solution is the product of these two solutions, since
the equation is of " degree " higher than the first.
Thus the solution is
(y—iix—CJ (y+3x-Cz)=o.
(dv\2
Example 25. — Solve the equation 5! -— ) — 8y8=o.
\ax/
3\ A v I *
Dividing by 5
Factorising \-~^- 1-26^ )(-/- — 1-265^) = °-
\CiX / \CLX /
Hence -^-+1-265^^ = 0 or -^-—1-265^ = 0.
Separating the variables and integrating
/^+i'265/^ = o or -%— 1-265 I dx =
J yl J J yi J
whence the complete solution is,
Two further examples are added to illustrate methods of solu-
tion other than those already indicated.
SIMPLE DIFFERENTIAL EQUATIONS 295
Example 26. — Solve the equation -j-^ — -^ — 14-^+2 AV = o.
dx3 dx* dx
The equation may be written (D3— D2— i.fD-f 24)^ = o
or (D-2)(D-3)(D+4)y = o,
whence y =
Example 27. — The equation j^= m*y occurs in the discussion of
the whirling of shafts. Solve this equation.
i. e., D4 = m*
or (D2— m2) (D2+m2) = o,
whence D= ±m or ±jm.
Hence y = a1emx-^-aze~mx-\-a3ejmx-\-ate-intx.
But e'x = cos x+j sin x, ejnue = cos mx+j sin mx
and ex — cosh AT+ sinh #,
i.e., enix = cosh w# + sinh mx.
y = ^ (cosh w#+ sinh m#) +«a(cosh mx— sinh w#)
+a3(cos w# +y sin ra#) +«4(cos mx—j sin w#)
= (aa+a4) cos wwr-f (a3— a4); sin w#+ («i+^2) cosh mx
+ («!— a2) sinh WAT
= A cos m# -f- B sin mx-\-C cosh m^+D sinh mx.
The constants A, B, C and D are found by consideration of the
conditions; four equations must be formed, these being found by
successive differentiation and by substituting for f- , -^ and ?? their
dx dxz dx3
values for various values of x.
Exercises 22. — On the Solution of Differential Equations.
1. If -p = 5#a— 2-4 and y = 1-68 when x = 2-29, find y in terms of x.
dzs ds
2. Given that — = 16-1 ; -=7 = 4-3 when t = 1-7 and s = 9-8 when
£££ £££
f = -2, find s in terms of t.
3. If ^ = 8y+5, find an expression for y.
296 MATHEMATICS FOR ENGINEERS
dv
4. Given that 8-76-^- +9* 15^ = 76-4 and also that ^ = 2-17 when
G/X
x = o, find an expression for y in terms of x.
5. A beam simply supported at its ends is loaded with a concentrated
load W at the centre. The bending moment M at a section distant x
from the centre is given by
™ W/ 1 \
M = — 4 -— x }.
2 \2 /
M dzv
If == = -j^, find the equation of the deflected form, y being the
deflection.
6. For the case of a fixed beam uniformly loaded, M, the bending
(w(P ,\ „) T, M d*y dy I
moment, =1-1 x2 1— K I . If ==- = — ^; -f- = o when x = - and also
1 2X4 / El dxz dx 2
when x = o ; and y = o when x = - , find an expression for y.
7. Solve the equation
dx i
the limits to T being Tx and T2, and to x being o and / ; the remaining
letters representing constants.
8. If -j- = 1- and v = o when x — s, find v in terms of x.
ax 4//x
9. If T! = the absolute temperature of the gases entering a tube of
length / and diam. D,
r2 = the absolute temperature of the gases leaving this tube,
6 = temperature of the water,
Q = amount of heat transmitted through the tube per sq. ft. per sec.
per degree difference of temperature on the two sides,
w = weight of gases along the tube per sec., and s = specific heat
of gases,
, dr QnT^dx
then — T H — — = o.
r— 0 ws
Find an expression for Q, x being the distance from one end of the
tube.
10. Find Q if Q^-nDdx-^-wsdr = o (the letters having the same
meanings as in No. 9, and the limits being the same).
11. If T-J and r2 are the inside and outside temperatures respectively
of a thick tube of internal radius rt and external radius rz, then
rfr H
dx ~~ 27rK I
I is the length of the tube, H is a quantity of heat, and the limits to
are o and rz—rv Find an expression for H, K being a constant.
SIMPLE DIFFERENTIAL EQUATIONS 297
12. A compound pendulum swings through small axes. If I =
moment of inertia about the point of suspension, h = the distance of
C. of G. from point of suspension, then
I X angular acceleration f i. e., 1^) = —mhd.
Find an expression for 6.
If p. = couple for unit angle — mh, prove that t, the time of a com-
plete oscillation, = 2/r \/ — (in Engineers' units).
r O
13. To find expressions for the stresses p and q (hoop) in thick
cylinders it is necessary to solve the equation
Solve this equation for p.
14. For a thick spherical shell, Up — radial pressure,
Find an expression for p, a being a constant.
15. If —~K.ttvdp = ~K.ppdv, prove that pv = constant, y being the
,. the specific heat at constant pressure .. KD
ratio -rr — - — r= — r— - of a gas, i. e., ==?•
the specific heat at constant volume KB
16. Solve for z in the equation
dz . w w2
3" H --- 7- zx — °-
dx g f
17. Solve the equation
dzy dy
^-i7^
and thence the equation
dzy dy
d^-l7Tx
dzs
18. Solve the equation — — 875.
at
dzs
19. Solve the equation ^73+875 = o.
(tt
20. Find the time that elapses whilst an electric condenser of
capacity K discharges through a constant resistance R, the potential
difference at the start being t^ and at the end vz, being given that
— K x rate of change of potential » = = •
jf*
21. If V = RC+L-r- and V = o, find an expression for C; C0 being
the initial current, i. e., the value of C when t — o.
J/~*
22. If V = RC-f L— -- , and V = V0 sin qt, find an expression for C.
ctt
298 MATHEMATICS FOR ENGINEERS
23. If -y- = -Vyz+2ay, find x in terms of y, a being a constant.
dx a •
24. An equation occurring when considering the motion of the
piston of an indicator is
ffix _a^ pa.
M^ + SM M
Solve this equation for x ; M, a, S and p being constants.
25. If -Py = Eig.
(an equation referring to the bending of struts), find y; given that
x = o when y — o, and v = Y when x = — •
26. To find the time t of the recoil of a gun, it was necessary to
solve the equation — = n Vxz— a2.
dt
If a = 47-5, n — 3-275 and the limits to x are o and 57-5, find t.
27. Solve the equation -^-\-2/~-\-nzx — o.
(Ait (Jit
dx
Take w2 = 200, /= 7*485 ; also let x = o and — = 10 when t = o.
Cut
28. The equation ^2"+2/^i +nzx = a sin qt
(It Q/v
expresses the forced vibration of a system. If n2 = 49, / = 3, q = 5, find
an expression for x.
dzV r
29. Solve the equation -,-=• — V — = o.
dx2 rz
30. If H is the amount of heat given to a gas, p is its pressure and
v its volume (of i lb.), then -,~ = 7— -Av^+np). Assuming that
dv (n— i) V dv */
there is no change of heat (i.e., the expansion is adiabatic and -5— = o J,
find a simple equation to express the connection between the pressure
and volume during this expansion.
31. Newton's law of cooling may be expressed by the equation
5" *<«-«
where k is a constant, and 6a is the temperature of the air.
If Q = 0Q when t = o, find an expression for 6.
32. The equation 1-04-^ +12-3— +13^—634=0
(It Ctt
occurred in an investigation to find 6, the angle of incidence of the main
planes of an aeroplane.
If t = o when 6 = i and — = o when t = o, find an expression for 6.
at
SIMPLE DIFFERENTIAL EQUATIONS 299
33. A circular shaft weighing p Ibs. per ft. rotates at o> radians per
second, and is subjected to an endlong compressive force F. The
deflection y can be found from the equation
d*y IF dfy _ p u
d^ + mfa* #
Solve this equation for y.
34. An equation relating to the theory of the stability of an aero-
plane is
dv
— = g cos a— kv
Civ
where v is a velocity ; g, a and k being constants. Find an expression
for the velocity, if it is known that v = o when t — o.
CHAPTER X
APPLICATIONS OF THE CALCULUS
THE idea of this chapter is to illustrate the use of the Calculus
as applied to many Engineering problems ; and the reader is
supposed to be acquainted with the technical principle involved.
The various cases will be dealt with as though examples.
Examples in Thermodynamics.
Example i. — To prove that (V— w) = — =5-, an equation occurring
T GtfT
in Thermodynamics,
where L = latent heat at absolute temperature T,
V = vol. of i Ib. of steam at absolute temperature T,
w = vol. of i Ib. of water = -016 cu. ft.,
P = pressure.
A quantity q of heat taken in at r-f-8r and discharged at r will,
according to the Carnot cycle, give out work = qr r - or approxi-
T-j-OT
4. 1 8f
mately q — .
T
Hence for i Ib. of steam at the boiling temperature,
Sr 87-
work = q — = L — ,
T T
but the work done = volume of steam in the cylinder x change in pressure
Hence (V-w)8P
and thus V— w=— r^,
T oir
or, as Sr becomes infinitely small,
X7. L,dr
V—w= —
T dP
Now -Tp is the slope of the pressure temperature curve (plotted from
300
APPLICATIONS OF THE CALCULUS
301
the tables) and can be easily found for any temperature r. Hence V can
also be found.
A numerical example will illustrate further.
It is required to find the volume of i Ib. of dry steam at 228°F.,
i. e., at 20 Ibs. per sq. in. pressure.
From the /When P= 19, * = 225-3, r= 460+225-3 = 685-3.
x 1.1 -^ .. P = 20,* = 228, T = 688.
steam tables ' .. -
^ „ P = 2i, t= 230-6, T = 690-6.
Plotting these temperatures to a base of pressures, we find that the
portion of the curve dealt with in this range is practically straight.
r
69O
689
688
687
686
685
/
3
f
/
/
2-65
t
s
i
—
H
!
s
/*
/
S
7
/9 eo p e»
FIG. 103. — Problem in Thermodynamics.
The slope of this line = 2-65 (Fig. 103), and this is the value of
-==, P being given in Ibs. per sq. in. and the latent heat in thermal
units. To change the formula to agree with these units,
778L dr
v =
Also L at 228° F. = 953.
V = -oi6-f
dP'
778x953x2-65
144x688
= 19-82 cu. ft.
Example 2. — To prove that the specific heat of saturated steam
(expanding dry) is negative.
Let
Q = the quantity of heat added.
H = total heat from 32° F.
I = internal energy of the steam.
302
Then
i.e.,
or
Now
Then
and
MATHEMATICS FOR ENGINEERS
H = internal + external energy
I = H-PV
81 = 8H-8(PV)=SH-(P8V+VSP).
= SH-PSV-V8P+PSV
= 8H-V8P
<\T r -*-* <\
8r
L
r
L
(from. Example i, neglecting^
\ w, which is very small J *
H =
dT^'305
Now the specific heat = heat to raise the temperature i°
hence
L.
= -305 - -
and since - is greater than -305, 5 is a negative quantity.
E. g., if t = 300° F., i. e., r = 761° F. absol.,
= 1115 — 210 = 905.
9°5
.'. s = '305 — ^-^
3 J 761
= - -882.
Work Done in the Expansion of a Gas.
Example 3. — Find the work done in the expansion of a gas from
volume vl to volume va.
There are two distinct cases, which must be treated separately; but
for both cases the work done in the expansion is measured by the area
ABCD = % areas of strips like MN (Fig. 104)
= ^,fp &v or I p dv more exactly.
J Vl
Case (a), for which the law of the expansion is pv = C.
/'»2 /"»2 / \"2
done = / pdv= I Cv~*dv = Cl log v }
Jit* J Vi \ h Jvi
Work
or C log r
r being the ratio of expansion, and = — .
Thus the work done = pv log r.
APPLICATIONS OF THE CALCULUS
303
Case (b), for which the law of the expansion is pvn — C, n having any
value other than i .
•1 **• 1
M
FIG. 104.
Work Done in the Expansion of a Gas.
V
FIG. 105.
Work done = / 2pdv = \ *Cv~ndv
J t7j • Dj
i— n
i
i—n
i
— (p»(
i — n
Pzvz — Pi
i—n
Work Done in a Complete Theoretical Cycle.
Example 4. — Find the work done in the complete cycle represented
by the diagram FGAB in Fig. 105.
The work done = area GABF = ABCD+GADH— FBCH
i—n
304 MATHEMATICS FOR ENGINEERS
Note that, if n = ^|
16
work done ==
If the expansion is adiabatic, and n is calculated according to
Zeuner's rule, n =
(q being the initial dryness fraction).
If q = i, then n = i-O35+-i = i-i35, so that the work done
To Find the Entropy of Water at Absolute Temperature r.
Example 5. — When a substance takes in or rejects heat (at tempera-
ture T) the change in entropy 8$ = — (8q = heat taken in) .
Let a = specific heat,
then o-Sr = 8q.
dq
Change in entropy from r0 to T = f —
JTT
dr
.
To
r_ - /i.e., in the change from\
For steam, the heat taken in at T = L { ).
\_ \ the liquid to the gas /
Hence the change of entropy = —
Efficiency of an Engine working on the Rankine Cycle.
Example 6. — Find the efficiency of an engine working on the
Rankine cycle ; using the T(f> diagram for the calculation.
Work done = area of ABCD (Fig. 106.)
= ABCK+ADMN-DKMN
= * x (TJ— T2)+heat taken in from
TI Tjtorj— (rtx'DK).
Now DK = the change in entropy from water at TZ to
water at TX
= loge — as proved above.
APPLICATIONS OF THE CALCULUS 305
Hence the work done = ^-^ (TI— r2) + (r1— T2) — ra log —
The heat put in
and thus the efficiency TJ =
r
r
being the dryness
fraction at
FIG. 106. FIG. 107.
Efficiency of an Engine.
Efficiency of an Engine working on the Rankine Cycle, with steam
kept saturated by jacket steam.
Example 7. — Find the efficiency of the engine whose cycle is given
by abcf in Fig. 107.
Work done = area abcf
_ f'aLj /the summation being of\
~~ j TI T \ horizontal strips )
[for L = 1115— -7/1
dr =1437— 7r
= a+br
= alog«-a+6(r2— TJ)
Total heat received = L2+r2— rj
total heat rejected = Lj
Hence the work done = (i) — (z).
. . (i)
(2)
306 MATHEMATICS FOR ENGINEERS
from which H,- = a log— +6(r2— TJ} — (L2— Lx) — (r2
Ti
= a log -+6(T2— TX) — (a+&T2— a— 6
-
Tl
Now the total heat received = L8+ra—
= L2+ra—
Hence
where a = 1437 and b = — "j.
Example 8. — To prove that the equation for adiabatic expansion of
air is pvy = C, where
_ specific heat at constant pressure _ Kp
' specific heat at constant volume Kw
Dealing throughout with i Ib. of air, let the air expand under
constant pressure from conditions p1 v1 rx to p^ v r.
Then the heat added = Kp(r— rx) = K/^- — ^~
Now keep the volume constant at v, and subtract as much heat as
was previously added : then the pressure falls to pz and the tem-
perature to ra.
The heat subtracted = K,(i— T,) = Kv(^—-
Now, if the changes are regarded as being very small, we may write
for v— vl and 8p for pi—pz
and thus — Kvv&p = Kpp8v
, (dp Kp rdv
whence / -~ = — ~ / —
/ P KJ i;
log p = — y log w + l°g (constant)
i. e., /> = Cy-v
or v~i = C.
APPLICATIONS OF THE CALCULUS 307
Examples relating to Loaded Beams.
Example g. — Prove the most important rule
M_E_ d*y
I R~ dxz
applied to a loaded beam ; M, I and E having their usual meanings,
and R being the radius of curvature of the bent beam.
Assuming the beam to be originally straight, take a section of length
/ along the neutral lamina, and let l-\-8l be the strained length at
distance y (Fig. 108).
A -
li\
FIG. 108. — Problem on Loaded Beam.
Then, if R = radius of curvature,
1+81 _
whence
or
but
and thus
I
R
1 + 81
R
I ~R
stress / / /R
1-4 - - _ — . J — • "i. _ - — . **_ _
strain 3/. y y
I R
R
M /
but it has already been proved (see p. 239) that y = — •
Hence
M = E
i ~R'
308
MATHEMATICS FOR ENGINEERS
The total curvature of an arc of a curve is the angle through which
the tangent turns as its point of contact moves from one end of the
arc to the other ; and the mean curvature is given by the total curva-
ture divided by the length of arc.
In Fig. 108 8$ = total curvature for the arc 65, and the mean curva-
S</>
ture = -5--
8s
We know that the slope of the tangent is given by ~
flwv
.-. tand> =
dy
-f-
dx
Now
, ,
tan $ = sec2 $ and
,
ds
2 sk ^^ ^ i^y
• • SCC O • ~^ ~^ ~5~ I ~~5
tis as \a^
i d (dy\
sec2 0 X s = x I T^ )
R as Va^r/
dx\dx) ds
dx* ds
Hence ^ = -^~ X -£ x cos2
R dx* ds
d*y
dx*
d<b
X --=
ds
dy
tan (b = -—•
dx
When, as for a beam, -~ is very small, (-*- j may be neglected in
comparison with i, and hence
This result may be arrived at more briefly, but approximately, in
the following manner : — •
80 = 8 tan 0 very nearly (when the angle is very small) .
Hence ~- = -- -- — — = — - tan <6 = rate of change of the tangent (for
8s 8s 8x
PM and PQ are sensibly alike).
APPLICATIONS OF THE CALCULUS 309
Thus ^^A.^^^y
ds dx dx dx*
i d2y
and _ = _£.
R dxz
M=/=E_F^
I y R dx*
In the use of this rule there should be no difficulty in finding
expressions for y in terms of x in
cases in which the beam is simply
supported; for an expression is
found for the bending moment at
distance x from one end, or the
centre, whichever may be more con- pi
venient, and then the relation
M d2y
•=- = E -n FIG. 109. — Beam Uniformly
n v& T j j
w;t Loaded,
is used ; whence double integration
from the equation so formed gives an expression for the deflected
form.
A few harder cases are here treated, the beam not being simply
supported.
Example 10. — A beam is fixed at one end and supported at the
other ; the loading is uniform, w being the intensity. Find the equation
of the deflected form.
We must first find the force P (part of the couple keeping the end
fixed) and then combine this force with the reaction at B calculated
on the assumption that the beam is simply supported. Referring to
Fig. 109: —
If the beam is simply supported, the bending moment at distance x
t -n Wl WX2
from B = — • x
2 2
Hence the actual bending moment
,, wlx wx2 _
= M = Px
2 2
-~~d2y wlx wx2 .._.
i.e., El ^-4= Px
dx* 2 2
whence, by integration,
_T dy wlx2 wx3 Px2 . ~
EI-j^- = hQ
dx 4 62
3io
MATHEMATICS FOR ENGINEERS
but
-f- = o when x = I
ax
for the deflected form is horizontal at this end.
wl3 wl3 PI* ,
o = f-C,
4 62
r _Pl2 wl3
1 ~~ 2 12
-rfrdy _ wlxz wx3 Pxz.Plz wl3
dx 4 6 2 2 12
Hence
T , , . „_ ze>#4 P#3 , Pl2x wl3x . „
Integrating, Ely = — — ^-C2.
12 24 6.2 12
In this equation there are the two unknowns P and C2, and to
evaluate them we must form two equations from the statements
y = o when x = o and y = o when x = 1.
If y = o when x = o, then it is readily seen that Ca = o.
wl* wl* PI3 , P/3 wl*
Also if x = l o = ----- ? — ----
12 24 6 2 12
1, T, Wl
whence P = -5-
o
Point' of Conhraflexure
/_
Fixed End
FIG. no. — Deflected Form of Beam.
If the beam were simply supported, the upward reaction would be
wl , ,, ,, , . wl wl 3 .
— , and thus the net reaction = 5- = ^wl.
2 288
Substituting -5- in place of P in the expression for y, we arrive at
o
the equation of the deflected form
the curve for which is shown in Fig. no.
We may now proceed to find where the maximum deflection occurs,
and also the position of the point of contraflexure.
dy w
dx 48EI '
dy
and
TX=°«
the solution of which, applicable to the present case, is x — -423^
APPLICATIONS OF THE CALCULUS 311
—•064— -423)
The maximum deflection is thus
wl*
•005 4 wl*
~EI~
To find C the point of inflexion or contraflexure
and 3-^ = o if x = o or if x = £/.
Example n. — A beam is fixed at one end and supported at the
other, the loading and the section both varying. Find the equation of
the deflected form.
Let m = bending moment at a point distant x from B if the beam were
simply supported, and let P= the force of the fixing couple. (See Fig. 109.)
Then M = m—Px
i. e., Elf^ = W_P*
, . dxz
.~dzy m Px
E^ = I-T
the equation being written in this form since I is now a variable.
By integrating E^= [X^dx-P F^dx+Cj. ...... (i)
t*x J 0 I J 0 1
dy
Now -j£ ss o when x = 1.
dx
i. e., Cj can be found, for the two integrals may be evaluated.
By integrating (i),
Ey = f f2(rf*)»-pr f \ (d^+c.x+c,.
J o .' o L JoJp1
But j/ = o when x = I and also when x = o, and thus C2 — o
and
J oJ o
i. e., P can be found.
The integrations must be performed graphically and with extreme
care, or otherwise very serious errors arise.
Example 12. — A beam is fixed at both ends and the loading and the
section both vary. Find the equation of the deflected form.
3i2 MATHEMATICS FOR ENGINEERS
Let Wj and m2 be the end fixing couples ; then to keep the system
in equilibrium it is necessary to introduce equal and opposite forces P
(Fig. in), i. e., P/+w2 = Wj.
Let m = the " simply supported " bending moment at section
distant x from the right-hand end.
Then M = m— mz— Px
dzv
and consequently El-r^ = m—mz—Px
d*m m Px
Integrating,
_
E
dy cxm, fxdx ^fxx
/- = / -=-dx—mtl -V-— P/ f
dx . o I *J 0 I J 0I
,4
-I —
FIG. in.
Now -jL*mQ when x = o or
dx
hence Cx = o (taking ^ = o)
and also, taking x = I,
Integrating again,
Ey = /• r^w-m. r r j w-p r ff w
J QJ 0 *• J OJ 0 i J 0-' 0 X
Now y = o when # = o and also when x = I.
Then taking x = o, C2 = o,
and taking # = /,
From equations (i) and (2) the values of wa and P (and hence Wj)
may be found, the integration being graphical (except in a few special
cases) ; and again it must be emphasised that the integration must be
performed most accurately.
APPLICATIONS OF THE CALCULUS
313
Example 13. — A uniform rectangular beam, fixed at its ends, is
20 ft. long, and has a load of 10 tons at its centre and one of 7 tons
at 5 ft. from one end. Find the fixing couples and the true B.M.
diagram.
This is a special case of Example 12 since the section, and therefore
I, is constant.
The B.M. diagram for the beam if simply supported would be as
ABCD (Fig. 112).
The bending moment diagram, due to the fixing couples only, would
have the form of a trapezoid, as APQD.
Unless the integration, explained in the previous example, is done
extremely carefully, there will be serious errors in the results ; and
since there are only the two loads to consider, it is rather easier to
FIG. 112. — Fixing Couples and B.M. Diagram of Loaded Beam.
work according to the Goodman scheme. [See Mechanics applied to
Engineering, by Goodman.]
According to this plan : (i) the opposing areas (i. e., of the free and
fixing bending moment diagrams) must be equal ; and (2) the centroids
of the opposing areas must be on the same vertical, i. e., their centroid
verticals must coincide.
To satisfy condition (i),
Area of ABCD =(^X5X5i-25)+(5I'25+67'5X5)+(^Xiox67-5)
= 762-5.
Area of APQD = - x 20 x (w1+m2) where m^ = AP\
and w = D
-> 10
Equating these areas, m1-{-mt — 76-25
(i)
314 MATHEMATICS FOR ENGINEERS
To satisfy condition (2), taking moments about AP,
r £
Moment of ABG = -X5I-25X-X5
= 427
Moment of BGH = -X5i-25X/5H — X5) = 854
r / 2 \
Moment of BHC = -x 67-5x1 5H — X5j =1405
Moment of DCH = — X 67-5 x ( io-| — x 10 )= 4500
2 V 3 /
i. e., total moment of ABCD about AP = 7186
, . _._ 20 20 200
Moment of APD = — x w, X — = n.
23 3
Moment of DPQ = — x mz X — = — m
Hence
(2)
200 .400
— ^1 + — m2~
The solution of equations (i) and (2) for m^ and w2 gives the results
mi = 44-71 and m2 = 31-54.
Thus PQ is the true base of the complete bending moment diagram,
AP being made equal to 44-71, and DQ equal to 31-54.
Shearing Stress in Beams.
Example 14. — To find an expression for the maximum intensity of
shearing stress over a beam section.
The shearing stress at any point in a vertical section of a beam is
always accompanied by shearing stress of equal intensity in a hori-
zontal plane through that point.
FIG. 113. — Shearing Stress in Beams.
FIG. 114.
We require to know the tangential or shearing stress / at E on the
plane CEC' (Fig. 113) ; this must be equal to the tangential stress in
the direction EF on the plane EF at right angles to the paper.
Suppose that the bending moment at CC'=M and that at DD'=M+SM.
Then the total pushing forces on DF > total pushing forces on CE, the
difference being the tangential forces on EFE/.
APPLICATIONS OF THE CALCULUS 315
Let P = the total pushing force on ECE'
,RC (stress = M "j
then P = / (stress) x area 1 y IV
J RE I area = bdyj
RE J-
M /"RC M
= v / by dy = •=- x ist moment of area ECE'.
I. 'RE
Now the tangential force on EFE' = stress x area
=/xEE'x8*
and this must equal the difference in the total pushing forces on DF
and CE, i. e., 8P.
Hence 8P = /x EE' x bx
i. e., -=r- x ist moment of area ECE'=/ x EE' ;
but — = rate of change of B.M.=shear=F (say).
Hence the maximum intensity of shearing stress/
~F T
= ist moment of area ECE'x ? X:
I "BE*
or, as it is usually written,
~ bl
where S = an area such as CEE' and y = distance of its centroid from
the neutral axis.
Example 15. — Find the maximum intensity of shearing stress, when
the section is circular, of radius r (see Fig. 114).
For this section
I = V.
4
Applying the rule proved above : —
F ,-r since zr corresponds to EE'
the maximum intensity = -^ J by dy in Example ^
du ('where u = r*—yz
dy~~
= - X mean intensity.
316 MATHEMATICS FOR ENGINEERS
Example 16. — A uniformly tapered cantilever of circular cross
section is built in at one end and is loaded at the other. The diam.
at the loaded end is D ins. and the taper is t ins. per in. of length.
Find an expression for the distance of the most highly stressed section
from the free end of the beam due to bending moment only. Neglect
the weight of the cantilever.
Let / be the length in ins. and W the load at the free end. Consider
a section distant x ins. from the free end ; then the diam. here is
D— tx, and the bending moment W#.
Also the value of I for the section considered is
. _
My /D— tx\ 64
Hence the skin stress f = -^- = Vfx( - IX -7-^-—-.
I V 2 / 7r(D— tx)
K _ 32W
D3- F ' ~
x (a constant),
and / is a maximum when the denominator is a minimum.
Let N = denominator
then =_
dx
dN
and -j— = o when 2t3x3—^Dt2x2-\-D3 — o
d 'X
i. e., when 2t3x3— zl)t*x2— D/a^2+D8= o
2t*xz(tx-~D)-I)(t2.v2-~Dz) = o
(te-D)(2^«-Dte-D») = o
(tx—T>)(2tx+T>)(tx—'D) = o
D D
t. e., when x — — or --- .
t it
Thus the stress is maximum at a section distant — ins. from the
t
free end.
Example 17. — To find the deflection of the muzzle of a gun.
This is an instructive example on the determination of the deflection
of a cantilever whose section varies.
The muzzle is divided into a number of elementary discs, the
volumes of these found (and hence the weights) so that the curve of
loads can be plotted. Integration of this curve gives the curve of
shear, and integration of the curve of shear gives the B.M. diagram.
APPLICATIONS OF THE CALCULUS
317
The values of I must next be calculated for each disc, and a new
curve plotted with ordinates equal to • : then double integration of
this gives the deflected form.
It is necessary to use the ordinates of this curve to find, first the
time of the fundamental oscillation and thence the upward velocity
due to the deflection.
Call the deflection at any section y, and the load or weight of the
small disc w.
Find the sum of all products like wy? and also find the sum of all
products like wy.
(Suitable tabulation will facilitate matters.)
Then T (time of oscillation)
If Y = maximum deflection, assuming the motion to be S.H.M.
then the upward velocity v is obtained from
vT = 27rY
or
v =
27rY
. Examples on Applied Electricity.
Example 18. — Arrange n electric cells partly in series and partly in
parallel to obtain the maximum current
from them through an external resistance R.
(Let the internal resistance of each cell = r,
and let the E.M.F. of each cell = v.)
Suppose the mixed circuit is as shown in
Fig. 115, i. e., with x cells per row and there-
7t
fore - rows.
x
Then the total E.M.F. of i row = ;n;
and total internal resistance of i row = xr,
¥1
but as there are — rows, the total internal
x
•I-!' I'l
H
FIG. 115. — Maximum Cur-
rent from Electric Cells.
resistance is - that of i row, i. e., the total
x
rx*
internal resistance = — ; but the E.M.F. is
n
unaltered : in reality the effect being that of one large cell, the area
being greater and thus the resistance less.
3i8 MATHEMATICS FOR ENGINEERS
total E.M.F. xv
Hence the current C =
total resistance rx
n
v
** , R"
n x
and C is maximum when the denominator is a minimum.
Let D = the denominator, then
dD r R , dD rx*
-=— = --- „ and -r- — o when R = — .
ax n xz dx n
i. e., external = internal resistance.
Example 19. — To find an expression for the time of discharge of an
electric condenser of capacity K, discharging through a constant
resistance R.
Let v = potential difference between the coatings at any time t.
v
Then, by Ohm's law, the current C = = .
But the current is given by the rate of diminution of the quantity
q and q = Kv.
dq dKv dv
Hence C = -^ = -^-= -K-
v Tjrdv
and thus ^ = — K-j- •
K at
If Vx = the difference of potential at the start, i. e., at / = o,
and V2 = the difference of potential at the end of T sees,
Separating the variables and integrating,
whence log =
Vt i
or =J = e KR
* 2
V2 _i
«. e., ^~ = e KR
or the time taken to lower the voltage from
Vjto V2 =
Example 20. — If R = the electric resistance of a circuit, L = its
self -inductance, C = the current flowing and V = the voltage, then
V = RC+L-
APPLICATIONS OF THE CALCULUS 319
Solve this equation for the cases when V = o, C = C0 sin qt and
For the case of a steady current V = RC since L is zero, and this
corresponds to the equation of uniform motion in mechanics, whilst
•j/-\
the equation V = RC+L-j- may be compared with that for accelerated
dt
motion, V being the force. Thus the second term of the equation may
be regarded as one expressing the " inertia " or " reluctance to change,"
and since the current may vary according to various laws, the rate of
change — can have a variety of values.
Clt
Dealing with the cases suggested : —
(i) V = o, then o = RC+L^-
dt
RC=~Lf
and this equation is solved by separating the variables and integrating.
Th ldC
Thus
•p j
whence — = loge C+loge A = loge AC
JLrf
_Rt
and AC — e L, A being a constant.
(2) If C = C0 sin 9*
j/-*
then — = qC0 cos qt.
dt
V = RC0 sin ^+L^C0cos qt
= C0VR2+L V sin U/+tan -1 -
then V0 sin qt = RC+L —
dl
i. e., V0sin qt = (R+LD)C
where' D= — .
at
Hence c V^ingf
and, using Theorem 2 of p. 288,
C = . V° = s
VR2+LV /
dC - —
To this must be added the solution of o = RC+L— , viz., C= Ke L ;
at
hence the complete solution is
V
320 MATHEMATICS FOR ENGINEERS
Example 21. — To find expressions for the potential and the current
at any points along a long uniform conductor.
At a distance x from the " home " end let the steady potential to
the ground = E and the steady current be i.
Let the resistance of i unit of length of the conductor — r and
let the leakage of i unit of length of the insulation = /.
Consider a small length of conductor 8x.
Its resistance = r8x and the leakage = I8x.
Hence the drop in potential —ixr8x (i)
and the drop in the current = E x I8x (2)
i. e., from (i) 8E = — ir8x
— - - '
and from (2) oi = — El8x
or -^ = -El (4)
Writing (3) and (4) in their limiting forms
dE _ _ . di _
dx dx
„.,- d2E d . . . di , .
Differentiating, TT= j~\tri ~ ~y^T ~ r ^ '5'
ct,\ d% dx
and -^r- r= -7-(— E/) = — /-j— = ril (6)
dxz dx* dx
To solve these equations, let D — ,
ax
then D2E = rlE
i. e., D2 = rl
D = ± Vrl.
dx dx
Separating the variables,
dE .- dE , -
— = Vrldx or — = — Vrldx.
Integrating,
log E = Vrlx+Ci or log E = — X/y/tf+Cg
or, if the constants are suitably chosen,
E = A cosh Vrlx+'B sinh Vrlx.
APPLICATIONS OF THE CALCULUS
321
In like manner, i = C cosh Vrl x+~D sinh Vrl x.
When x = L
E = A cosh Vr/L+B sinh Vr/L.
When x = o
E = A
and hence the constants can be found.
Examples on Strengths of Materials.
Example 22. — To find the shape assumed by a chain loaded with its
own weight only ; the weight per foot being w. To find also expres-
sions for the length of arc and the tension at any point.
Let s = the length of the arc AB (Fig. 116) : then the weight of
this portion = ws.
\IVS
7
FIG. 116.
FIG. 117.
Draw the triangle of forces for the three forces T, T0 and ws
(Fig. 117).
Let it be assumed that T0 (the horizontal tension) — we, where c is
some constant.
Then, from Figs. 116 and 117
dy _
dx =
Now, as proved on p. 201,
.. ws ws s
, = tan 6 = =- = — = -•
ax TO we c
ds
dx*
sds
322 MATHEMATICS FOR ENGINEERS
To integrate the left-hand side, let u = c2+s2
du
then r = 2s
ds
( sds rsdu
and / -7= = = / 1 = u* = V c2 -1- s2
./ Vc2 + S2 J 2SU$
Thus, by integration of equation (i),
Now at the point A (Fig. 116) 5 = 0 and y = o
hence Vcz = Cj or Cx = c.
Thus Vc2+s2 = y + c.
Squaring c2+s2 = y*-\-cz+2yc
or sz = yz+2yc and s — Vyz+2yc
s dy dy
but as proved above -= -f- or s — c^-
c ax dx
dv /— =
hence c-^-= V yz -\-2yc.
Separating the variables
dy dx
_ __ ^_
e" —c2 ~ c
r dy [
Integrating / /. , ., =± = /
J V+c2— c2 J
/. , .
V(y+c)— c c
and this integral is of the type discussed on p. 151 ; the result being
c J c
Now x — o when y = o, x being measured from the vertical axis
through A, and thus logf- j = C2 or C2 = o.
Thus f=
or in the exponential form
X
cec = y-\-c-\- Vyz+ zyc.
Isolating the surd cec — (y+c) = Vy*+2yc.
2x x
Squaring cze c +y2+c2+2yc— 2(y-\-c) ce* = y*+2yc
2x x_
or czec —-2.ee* (y+c)+c2 = o.
APPLICATIONS OF THE CALCULUS
323
Dividing through by cec
cec +ce c — 2 (y-\-c),
i.e., (y+c)=C-(e°+e~°).
If now the axis of x be shifted downwards a distance c, then the
c I X' — ~\ x
new ordinate Y = y-\-c and Y — - \ec+e c ) — c cosh -•
v: Scale
FIG. 1 1 8. — Catenary Form of a Cable.
Again, since Y = y +c =
dx
dx
dx
and also
Then
d ,x c . , x . .x
^-c cosh - = - sinh - = smh -•
ax cec c
dy .X
•—• = sinh -
dx c
but it has already been proved that
dy_ s
henee
dx c
s . , x
-=smh- or s =
c c
324 MATHEMATICS FOR ENGINEERS
To find the tension T at any point
T2 == wzsz+wzcz from Fig. 117
= wz(sz+cz)
= wz (c+y)z = wzY2
or T = wY.
Thus the form taken by the chain is that for which the equation is
fx\
._^ Y = ccosh(-J, the equation of the catenary: the
length of arc is given by 5 = c sinh -, and the tension
c
at any point is measured by the product of the
ordinate at that point and the weight per foot of the
chain.
Fig. 118 shows the catenary for a cable weighing
3-5 Ibs. per foot and strained to a tension of 40 Ibs.
weight, and the method of calculation for the con-
struction of this curve is explained on p. 358 of PartT.
The tension at 10 ft. from the centre = 3-5 x 15-9
FIG. 119. = 55-6 Ibs. weight, since the ordinate there is 15-9.
Example 23. — To find the time of oscillation of a compound pendu-
lum swinging through small arcs.
Let I be the moment of inertia of the pendulum about an axis
through the point of suspension O (Fig. 119), and let h = the distance
of the C. of G. from the point of suspension.
Then the couple acting, to produce the angular acceleration,
= moment of inertia x angular acceleration.
[Compare the rule for linear motion, Force = mass x acceleration.]
Now the angular velocity = -j-
dt
and hence the angular acceleration
= <8» '
Thus the couple acting = 1-^
and this couple is opposed by one whose arm is h sin 6, as is seen in
the figure.
dzd
Thus *dt*~= — mhsvn.6 — — mhd
since 6 is supposed to be small, and consequently sin 6 = 6
or dzd _ mh . _ z ., 2 _ mh
dt*~~ ~~T~ " ~T~
and £+••*-«.
APPLICATIONS OF THE CALCULUS 325
This equation is of the type dealt with in Case (3), p. 283, and the
solution is 6 = A sin (<ef-f B).
2-7T
The period of this function is — ; also the couple for angular dis-
co
placement Q = mh6 ; hence the couple for unit angular displacement
(denoted by /*) = mh.
27T /T~ /I
Hence t = — = 2rr \/ — r = 2ir \f -
co ^ mh v p.
or / = 2n A / — if engineers' units are used.
k i*g
This might be written in the easily remembered form,
unit moment of inertia per unit twist'
If this formula is to be used in the determination of the modulus of
rigidity of a sample of wire by means of torsional oscillations, h must
be replaced by /, the length of the wire.
If /= skin stress, T «= torque, and C = modulus of rigidity, d — dia.
of wire; then
whence
but
and
0=g and T =
A 32T/
M ~ 6 ~ yl
t=2w \/ —
/ 32/1
hence C = — ==— - and thus C can be determined.
As regards the units, if / is in feet, I is in Ibs. ft.2, t in sees., and
d in feet,
feet x Ibs. x ft.2 x sec.2
then C =
feet x ft.4 x sec.2
Ibs
= ~ffT' **• e-> C *s m Ibs. per sq. foot.
If I is in Ibs. ins.2 and d is in ins., then C will be in Ibs. per sq. in.
Example 24. — To find formulae giving the radial and hoop stresses
in thick cylinders subjected to internal stress.
We may attack this problem by either of two methods : —
Method i. — In (a) Fig. 120 let the outside radius = rt and the inside
radius = r0 ; also let the internal pressure be p, and the crushing stress
at right angles to the radii, or the hoop stress, = q.
326
MATHEMATICS FOR ENGINEERS
It is rather easier to consider the stress on the outside to be greater
than that on the inside : thus for an annulus of radius r and thickness
$r, we take the internal stress as p and the external stress as p-\-8p.
Considering the element RS of the annule (subtending an angle of
80 at the centre), and dealing with the radial forces,
Total radial force = (p-\-8p) x outer arc— px inner arc
= (p+8p) X (r+8r)86-pr86
= (pr+p8r+r8p+8p.8r-pr)86 .
= (p8r+r8p+8p.8r) 86
(for a unit length of the cylinder) .
This is balanced by two forces each q 8r . 86, for
2 . 86 86
— — = sin - = — nearly
q8r 2 2
[(6) Fig. 120]
i. e., x = q8r.86
x being the radial force.
FIG. 1 20. — Stresses in Thick Cylinders.
Thus (p8r+r8p+8p.8r)86 = q8r.8d
or, when 8r is very small,
p dr-\-r dp = q dr.
Assume each longitudinal fibre to lengthen the same amount due to
the secondary strains.
Then if <r = Poisson's ratio and E = Young's modulus for the
material,
p i
the extension due to p will be ^ X -
and the extension due to q will be ^ X -
Si, (T
then, since the total extension is to be constant,
i. e.,
~r^ — constant,
(T-tL
p-\-q = 2A, say, for a and E are constants.
APPLICATIONS OF THE CALCULUS
Hence v dp -\-pdr = qdr
= (2 A— p}dr
i. e., rdp = 2(A—p)dr.
Separating the variables and integrating,
f dp _ fdr
J 2(A-p) J r
i.e., — | log (A— p) + log C = log r
C
327
or r =
(A-pfl
C
r
-r>
or p = A + -2
but
and hence
The constants A and B are found from the conditions stated in any
example.
Method 2. — According to this scheme q is taken as a tensile stress.
By the thin cylinder theory ; — consider the equilibrium of the half
elementary ring of unit length [(c) Fig. 120].
Then (pX2r) — (p+8p)2(r+8r) = 2q8r
whence qdr = —pdr—rdp.
From this point the work is as before except
that 2 A is written for p — q and not for p-\-q
as in Method, i.
Example 25. — To find expressions for the
stresses in Thick Spherical Shells.
FIG. 121.
Let p = the radial pressure, q = the hoop tension.
Take an elementary shell at radius r, the thickness being 8r (Fig. 121).
Then 7rrzp—n(r+8r)2(p+&p) =
When 8r is very small this equation reduces to
— 28r— r8 = 2q8r
and hence
2q = —
dp
-
328 MATHEMATICS FOR ENGINEERS
Assuming the volumetric strain to be the same everywhere,
where lv = the circumferential strain, and thus 2/2,= the superficial
strain, and lx = the radial strain,
then it follows that
^-- P.-C
<rE o-E"
i. e., 2(I—p — Constant = 3A (say).
Now 2<7=— 2*— r~
Separating the variables and integrating,
[-$L
J T
i. e., — log r = - log (A+£) -f log (^
whence
or
i _ _ 2B_A
Also zq •
3 . 2B
_B
Euler's Formulae for Loaded Struts.
Example 26. — To obtain a formula giving the buckling load for a
strut of length L and moment of inertia I.
Applying the ordinary rule
M =Ed*y
APPLICATIONS OF THE CALCULUS
Bending moment at Q = M = — Py [(a] Fig. 122]
329
i. e.,
Let
then
dx2
IE/'
IE
and the solution of this equation is, according to Case (3), p. 283,
y — A sin (
fW (c)
FIG. 122.
The various conditions of end fixing give rise to the following
solutions : —
Case of ends rounded. — When x — o, y = o
then o = A sin (o+B) and A is not zero
so that B = o.
When
i. e.,
x = — , y= Y [(a) Fig. 122]
xr . wL
Y = A sin —
2
Obviously Y is the amplitude, i. e., A = Y
0)L
or
Thus we may write
whence
and
i — sin
2
7T . 0)1
2 2
0>L = 7T
7~P _
V fEXL =
7T2IE
, L2 '
(- being the simplest angle
\2 \
having its sine = i )
330
MATHEMATICS FOR ENGINEERS
Case of both ends fixed. — -The form taken by the column is as at
(&) Fig. 122. The half-period of the curve is evidently — in this ease.
27T
But the period = —
27T
IE
whence
47T2IE
L2
Case of one end fixed. — The form taken by the column is as at
(c) Fig. 122. The half-period in this case is f L, but, as before proved,
. , . 27T
the period is given by —
O)
Hence -L — —
2co
Now
/*
" V IE
Q7T2IE
4P
97T2IE
4L2 '
L2 =
P =
Tension in Belt passing round a Pulley.
Example 27. — To compare the tensions Tx and T2 at the ends of a
belt passing round a pulley ; the coefficient of friction between the belt
and pulley being p., and the angle of lap being 6 radians.
T+ST
FIG. 123.
FIG. 124.
Consider a small element of belt subtending an angle of dd at the
centre of the pulley (see Fig. 123) : then the tensions at the ends are
respectively T-f-oT and T.
APPLICATIONS OF THE CALCULUS 331
Resolving the forces horizontally
s»a s/i
/T> I »T\ "" T> ™" T-»
(T+8T) cos -- Tcos — uP
2 2
f./\
i.e., 8T cos — = uP
2
or in the limit dT = /J? ......... (i)
80
for cos -- > cos o, z. 0., i.
2
Resolving the forces vertically,
Sj/1
P=(T+8T+T)sin~
' . 80 , ._ . 80
= 2T sin -- f-8T sin —
2 2
_, .
= 2T -- h8T— (for sin — = —
22 22
when the angle is small)
In the limit P = TdO .......... (2)
Then, combining equations (i) and (2),
dT = pTdS
Separating the variables, / -=- = p. / dd
J T2 i Jo
T
Integrating, log«-f^ = \iB
*-2
T, n8
or =*=• tr
*i
Friction in a Footstep Bearing.
Example 28. — To find the moment of the friction force in a footstep
bearing; the coefficient of friction being p., R = radius of journal and
W= total load.
(a) Assume that the pressure is uniform over the bottom surface,
i. e., W = 7rR2^>, where p is the intensity of the pressure.
Take an annulus at radius r, and of thickness 8r (Fig. 124).
Area of the annulus = 2nr8r
Pressure on the annulus = 2-nrdrp
Friction force on the annulus — -z-nrbrpp
and hence the moment of the friction force on the annulus
= 2irr 8r pp x r
332 MATHEMATICS FOR ENGINEERS
/R
Up 2nrzdr
R»
T
W
= -
i. e., the moment is the same as it would be if the whole load were
supposed concentrated at a distance of two-thirds of the radius from
the centre.
(&) Assume that the intensity of pressure varies inversely as the
velocity,
i. e., p= K X -
v
the velocity at radius r — vr =
so that p = K x
Then the pressure intensity on an annulus distant r from the centre
so that p = K x — = - (say).
r v
and the total pressure on the annulus
also the friction force = p X this pressure.
Hence the moment of the friction force on the annulus
and the total moment of the friction force
=v
Now the total load W = / intensity x area
J o
•/*
= I p X 2nrdr
J o
=r
Jo
APPLICATIONS OF THE CALCULUS 333
Hence the moment of the friction force
= 27rwR X —
2
-••Wxf
i.e., the effective radius is now £ and not f, as in Case (a).
Example 29. — To find an expression for the moment of the friction
force for a Schiele Pivot.
Assume that the pressure is the same all over the rubbing surface,
that the wear is uniform and that the normal wear is proportional to
the pressure p and to the speed v.
Referring to Fig. 125,
and thus
= normal wear oc pv, i. e.,
8n oc pr or &n = Kpr.
Let the tangent at the point P make the angle 6 with the axis, then
if t = length of tangent, t sin 6 = r.
Now 8h — vertical drop = - — ^
sintf
Also 8n = Kpr = Kpt sin 6
whence bh = Kpt.
Now 8h is constant, p and K are constant ;
hence t must be constant and the curve is that
known as a tractrix (i. e., the length of the
tangent from the axis to any point on the
curve is a constant).
To find the moment of the friction force : —
On a small element of surface, the friction force
= lirr 8s X p X p.
and the moment of the friction force
= 2nr8s fip x r.
Now 87 = 8s sin 6.
Hence the total moment of the friction force
dr
FIG. 125.
sin 6
tsinti
but
X
W
Hence the total moment of the friction force = /*\V/.
334
MATHEMATICS FOR ENGINEERS
Examples on Hydraulics.
Example 30. — To find the time to empty a tank, of area A sq. ft.,
through an orifice of area a sq. ft., the coefficient of discharge being Cd.
If the height of the water above the orifice at any time is h, then
the velocity of discharge = v = V^gh.
Hence the quantity per sec. = Cd av
and the quantity in time 8t — Cd av8t.
This flow will result in a lowering of the level in the tank by an
amount &h, so that the volume taken from the tank in time 8t = A 8h.
Hence A8h = Cd aV^ghSt.
Here we have a simple differential equation to solve, and separating
the variables and integrating
Adh where hz = initial height
Ax = final height
o A
FIG. 126. — Triangular Notch.
If A1 = o, then the time to completely empty the tank
Example 31. — To gauge the flow of water by measurements with a
triangular notch.
Let the height at the notch be H, and consider a small element of
width b, thickness 8h, and height above the apex of the notch (H— A).
b G
From Fig. 126, - = (H— h) tan - where 6 = the angle of the notch,
whence
6 =2(H— h) tan-.
APPLICATIONS OF THE CALCULUS 335
Now the area of the element = b 8h
and the velocity of water at that height = VzgX height =
A
Hence the actual quantity flowing = CdX2(H— h) tan -
and the total quantity flowing for the height H
• fi
= FzVzgCa tan-(H
J o
tan
/) /~n
-J ^H
1 /— i •
J5 2
a
~Li 6 = 90° (a common case), tan- = i,
Q
and then the discharge = — V2gCdH* = 2-66 H? if Cd= -62.
Example 32. — To estimate the friction on a wheel disc revolving in
a fluid.
Let the friction per sq. ft. —fvx and let the disc (of inside radius R2
and outside radius Rx) revolve at n revs, per sec.
The velocity of an annulus at radius r = 2nnr
and thus the friction force per sq. ft. on this annulus
= (2nnr)xf.
Hence the moment of the friction force on the annulus
= /(2 nnr)x X -Zirrbr X r
and the total moment of the friction force on one side of the disc = M
Rg
The total moment (i. e., on the two sides) = aM
and H.P. lost in friction = — — .
550
If x = 2
= 49.6/n»{R1»-Rf«}.
336 MATHEMATICS FOR ENGINEERS
Example 33. — To establish a general rule for determining the depth
of the Centre of Pressure of a section below the S.W.S.L. (still water
surface level).
Suppose the plate (representing a section) is placed as shown in
Fig. 127. Consider a small element of area 8a, distant x from OY, the
vertical distance being h.
Let X = distance of the C. of G. from OY,
X = distance of the C. of P. from OY,
and let H and H be the corresponding vertical distances.
Let P = total pressure and A = total area,
k = swing radius about OY,
and p — weight of i cu. ft. of water.
S.W.S.L.
IX
FIG. 127. — Centre of Pressure.
The whole pressure on the element
= intensity of pressure X area = ph X 8a
Whole pressure on surface = 2px sin a8a approx.
or fpx sin a da actually,
i.e., P = psinaxfxda
— p sin aX ist moment of area about OY
= p sin a X AX,
but X sin a = H.
To find the position of the C. of P., take moments about OY.
Then PxX = 2 moments of the pressures on the elements
= Sp sin a x8a X x
= psina2#2Sa approx.
= psinafxzda actually
= p sin a X and moment about OY
= p sin a X A&2.
APPLICATIONS OF THE CALCULUS 337
Now P = pHA
so that pHAX = psinaAA2,
i. e., HX = sin a . kz
but H = X sin a
hence XX = A2.
Thus if X is known, X can be calculated.
If the body is not symmetrical, then Y (the distance of the C. of P.
from OX) must be found by taking moments about OX.
In a great number of cases a = 90°, so that sin a = i, and thus
Example 34. — A triangular plate is placed with its base along the
S.W.S.L., the plate being vertical. Find the depth of the centre of
pressure below the surface.
For this section I about S.W.S.L. = — bh3
and thus k2 =
also H = -h. '
3
- k2 A2X3 h
Hence H = — = -^—j^ = -
H 6x& 2
Example 35. — A circular plate has its upper edge along the S.W.S.L
Find the depth of its Centre of Pressure below the S.W.S.L.
For a circle, Idiam. = 7- d*
64
and thus k2 about diam. =
4
and hence, by the parallel axis theorem,
A2 about S.W.S.L. = — H ~ ) = —
also H = -
•2.
= kz 16 ^d
Hence H = — = —5-= "-3-
H d
338
MATHEMATICS FOR ENGINEERS
Example 36. — Forced vortex (i. e., water in a tube rotated round a
vertical axis). To find the form taken by the surface of the water.
Let the rotation be at n R.P.S.
Consider an element at P (Fig. 128).
tan 6 =
th
tan 6 = the slope of the curve taken = -^
dn
vertical force _ weight *of particle
horizontal force centrifugal force on particle
_mxgr
FIG. 128. — Forced Vortex.
Separating the variables
Integrating
dh 4tr
4n2n2rdr = gdh.
^=gh.
A = ^xr*.
Now
is a constant, and thus
h = constant xr2, this being the equation of a parabola.
Hence the surface of the liquid will be that of a paraboloid of
revolution.
An Example from Surveying.
Example 37. — Prove that a cubic parabola is a suitable " transition "
curve.
In order that the full curvature of a railway curve may be
approached gradually, a curve known as a transition curve is inter-
posed between the straight and the curve. It must be so designed
that the radius of curvature varies inversely as the distance from the
starting point (on the straight, because there the radius is infinite) .
i d2y
As before proved,
dx2
or more exactly
ffiy
dx2
APPLICATIONS OF THE CALCULUS
339
For the cubic parabola we may assume an equation
y = px3,
where x is the distance from the straight, along a tangent, and y is the
offset there (obviously p must be very small) .
If y = px3, -r- = ipx2, ~ = 6px.
Hence = = -
as a first approximation, or — = 6px nearly (for p3 is very small) .
T T T K"
Hence
R
or K = — X - = —
6p x x
Roc -.
x
Exercises 23.
1. A cylindrical tank is kept full of water by a supply. Show that
the time required to discharge a quantity of water, equal to the capacity
of the tank, through an orifice in the bottom equals half the time
required to empty the tank when the supply is cut off.
A tank 10 ft. high and 6 ft. diam. is filled with water. Find the
theoretical time of discharge through an 8" diam. orifice in the bottom.
2. A tank empties through a long pipe discharging into the air. If
Kv2
the head lost in the pipe is written hi = - — , show that K can be found
from the expression,
where Ax is the level of the water in the tank at the time tt
and hz is the level of the water in the tank at the time t2
measured from the centre of the discharge end of the pipe.
A = area of cross section of the tank.
a = area of cross section of the pipe.
An experiment with a tank 15-6 sq. ft. in cross section and a 4" diam.
pipe gave the following results :
Time, t inins.
Level in tank, h.
0
I
2
38-35
32-84
3
4
23-19
I9.O2
Find the value of K for the pipe.
340 MATHEMATICS FOR ENGINEERS
3. Use the following table to obtain -= and thence find the volume
of i Ib. of steam at 160 Ibs. absolute pressure per sq. in.
Absolute press. (Ibs. per sq. in.)
159
1 60
161
Temperature (F°)
363-1
363-6 •
364-1
The latent heat of i Ib. of steam at 160 Ibs. per sq. in. pressure is
858-8 B.Th.U.
4. Find the " fixing moments " for a beam built in at its ends and
40 feet long, when it carries loads of 8 tons and 12 tons, acting 15 feet
and 30 feet respectively from one end.
5. A tank of constant cross section has two circular orifices, each
2" diani., in one of its vertical sides, one of which is 20 ft. above the
bottom of the tank and the other 8 ft.
Find the time required to lower the water from 30 ft. down to 15 ft.
above the bottom of the tank.
Cross section of the tank =12 sq. ft.
Coefficient of discharge = -62.
6. A hemispherical tank 12 ft. in diam. is emptied through a hole
8" diam. at the bottom. Assuming that the coefficient of discharge is
•6, find the time required to lower the level of the water surface from
6 ft. to 4 ft.
7. A vertical shaft having a conical bearing is g" in diam. and
carries a load of 3^ tons ; the angle of the cone is 120° and the co-
efficient of friction is -025. Find the horse power lost in friction when
the shaft is making 140 revolutions per minute.
Assume that the intensity of pressure is uniform.
8. A circular plate, 5 ft. diam., is immersed in water, its greatest
and least depths below the surface being 6 ft. and 3 ft. respectively ;
find
(a) the total pressure on one face of the plate,
(b) the position of the centre of pressure. •»<
9. An annular plate is submerged in water in such a position that
the minimum depth of immersion is 4 ft. and the maximum depth of
immersion is 8 ft. If the external diam. of the plate is 8 ft. and the
internal diam. 4 ft., determine the total pressure on one face of the
plate and the position of the centre of pressure.
10. One pound of steam at 100 Ibs. per sq. in. absol. (vol. = 4-45
cu. it.) is admitted to a cylinder and is then expanded to a ratio of 5,
according to the law pv1-06 = C ; it is then exhausted at constant
pressure.
Find the net work done on the piston.
11. Find the loss of head h in a length I of pipe the diameter of
which varies uniformly, being given that
„ 4fLv* 4Q
H= J , , and v = ^~.
•2gd Ti-d2
{Let diam. at distance x from entry end = rfe+K#
where de = diam. at entry}.
APPLICATIONS OF THE CALCULUS 341
12. Taking the friction of a brass surface in a fluid as -22 Ib. per
sq. ft. for a velocity of 10 f.p.s. and as proportional to w1'9, find the
horse power lo"st in friction on two sides of a brass disc 30" external
and 15* internal diam. running at 500 r.p.m.
13. A rectangular plate 2 ft. wide by 5 ft. deep is immersed in
water at an inclination of 4OC to the vertical. Find the depth of the
centre of pressure, if the top of the plate is 6 ft. below the level of
the water.
CHAPTER XI
HARMONIC ANALYSIS
Fourier's Theorem relates to periodic functions, ot which
many examples are found in both electrical and mechanical engi-
neering theory and practice : it states that any periodic function
can be expressed as the sum of a number of sine functions, of
different amplitudes, phases and periods. Thus, however irregular
the curve representing the function may be, so long as its ordinales
repeat themselves after the same interval of time or space, it is
possible to resolve it into a number of sine curves, the ordinates of
which when added together give the ordinates of the primitive
curve. This resolution of a curve into its component sine curves is
known as Harmonic Analysis ; and in view of its importance, the
simpler and most direct methods employed for the analysis are
here treated in great detail.
Expressed in mathematical symbols, Fourier's theorem reads
or y =
+Bj cos §tf+B2 cos 2qt+~B3 cos 3qt-\- . . .
the latter form being equivalent to the first, since
Aj sin gtf+Bj cos qt — B sin (qt-\-Cj)
provided that B and c± are suitably chosen.
For the purposes of the analysis the expression may appear
simpler if we write 6 in place of qt.
Thus
y = A0+A1sin^+A2sin2^+A3sin3<9+ . ..-.
+B1cos0+B2cos20+B3cos30-|- . . .
Of the various methods given, three are here selected and
explained, these being easy to understand and to apply.
Dealing with the three processes in turn, viz., (a) by calculation,
(b) by a graphical interpretation of method (a), and (c) by super-
position, we commence with the study of method (a).
342
HARMONIC ANALYSIS 343
Method (a): Analysis by Calculation. — Before actually pro-
ceeding to detail the scheme of working, it is well to verify the
following statements.
ft*
cos 0 = 0, this being self-evident, since the area under a
.'o
cosine curve is zero, provided that the full period is considered.
/**
I cos mO cos nOdO — o ............ (i)
.'o
for
cos mO cos nO = ^{cos (m-\-n)6— cos (m— n)0}
and hence
/•2ir
/•2ir r2n rZv
I cos m0cosn9d& = || cos(m + n)OdO— 1| cos(m—n)6d0
Jo Jo Jo
= o— o
(for both are cosine curves over the full period or a multiple of the
full period).
rZv f2jr /-2ir
I cos m6 sin nO dO = | sin (m+n)B dO—% sin (m—n)6dO
Jo J o Jo
= 0 .... ........ (2)
fZir ,fZir rZir
cosz6d6 =4 cos20dO + % dd
Jo Jo Jo
= 0 + £(27r-o)=7r ....... (3)
/•2ir f /-2ir /-2n- ^
sin mO sin M^ ^ = $\ cos (m—n)6 dd— cos (m+ri)0d6 \
Jo U o y o J
= |{o-o}-o ......... (4)
and
/•2ir /-2ir fZir
I sin2^^ =J ^— | cos20d0
Jo Jo Jo
= \\2ir— o} — O
= » ............ (5)
To proceed with the analysis : —
We are told that
y = A0+A1cos0+A2cos20+A3cos30+ ...
+Bj sin 6+B2 sin 20+B3 sin 3^+ • • •
and we wish to find the values of the coefficients A0, Aj . . .,
Bi, B2, etc.
344 MATHEMATICS FOR ENGINEERS
If we integrate throughout (with the limits o and 2ir), every
term on the right-hand side, except the first, will vanish,
i.e.,
r2w f
yd9 = A,
Jo J
/•2ir
or I ydO — A0X (2-*— o)
3 o
I P*
whence A0 = - —I y dO
= the mean value of y (Cf . p. 183)
so that A0 is found by averaging the ordinates ; but in the majority
of cases an inspection will show that A0 is zero.
To find A1 : — multiply all through its coefficient, viz., cos0, and
integrate, then
fZir r'2ir rZir rZ
I ycos$dO= I A0cosOdO-\- I A1cos20^+|
Jo ./o Jo ,' o
rZn _ rZir
Jo lC( Jo
f2
I y cos 6 dO = o+TrAj+o+o . . .
Jo
or
J o "
+0+0 . . . [from (3), (2) and (i)]
2 f£jr
whence Ax = / y cos 6 dO
= twice the mean value of (y cos 6}
i. e., a certain number of values of y must be taken, each being
multiplied by the cosine of the angle for which y is the ordinate,
the average of these found, and the result multiplied by 2.
The values of A2, A3, etc., may be found in like manner by
multiplying through by cos 26, cos 3$, etc., in order, and performing
the integration as above.
To find B! : — multiply throughout by its coefficient, viz., sin 6,
and integrate, then
/2ir [Z* ,-Zir fZir
ysinBdO = I A0sm@d@-\- I A,sin0cos0^+ / A
o J o J o Jo
,"Zir _ rZir
+JoBlS1 +JoB2
= o+7rBj [from (2), (4) and (5)]
_, 2 r27r
•'• Bj = 1 y sin 0 dQ = 2 X mean value of (y X sin 6}
so that the values of B,. B2, B3, etc., may be found
HARMONIC ANALYSIS
345
Actually, the values of the coefficients Av A2, Br, B2, etc., are
found by dividing the base into ten or. eight divisions and averaging
the mid-ordinates for these divisions. To determine the absolute
values, an infinite number of ordinates should be taken, but this
would of course be quite out of the question as far as an ordinary
calculation is concerned.
The work is made clearer by suitable tabulation, as will be seen
from the following example.
Example i. — Resolve the curve ABCD (Fig. 129) into its component
curves : it being understood that no higher harmonic than the first
occurs.
e
O -36° 72° K>8 144
FIG. 129. — Harmonic Analysis.
[The term containing 6 is spoken of as the fundamental, and that
containing 26 as the first harmonic.]
Thus y = A0+B sin (6+cJ +C sin
or y= A0-|-A1cos0+A2cos20+B1sm0+B2sin20
will represent the function in this case.
A glance at the figure will show that the curve is symmetrical about
the axis of 6 ; thus we observe that the average ordinate = o, or
Aft = o.
346
MATHEMATICS FOR ENGINEERS
Divide the base into 10 equal parts, erect the mid-ordinates and
tabulate the values as follows : —
(a) (/•) (c) (d) (e) (/) (g)
Ordinate No.
9
1
sin 0
cos e
sin 26
cos 26
I
1-56
18°
•309
•951
•588
•809
2
375
54°
•809
•588
•95 1
-•309
3
4
90°
I-
o
o
— I
4
2-91
126°
•809
—588
—951
-•309
5
1-13
162°
•309
-•951
—588
•809
6
-1-13
198°
— •309
-•951
•588
•809
7
— 2-91
234°
-•809
—588
•95 1
— 309
8
-4
270°
— I
o
0
T
9
-3'75
306°
— •809
•588
—951
-•309
10
-1-56
342°
309
•951
—588
•809
Then A0 = mean value of y = o.
A! = 2 X mean value of (y cos 6} .
To obtain the values of y cos 6, corresponding figures in columns
(6) and (e) must be multiplied.
Then
10 1 +2-91—3-75)
= o
-9i\
)
Similarly, Bx = — X sum of the products of columns (&) and (d)
io
i-i3 + i-56) + 1(4+4) \
9i + 2-9i+3-75) /
= -2x20-43=4-086
2
A2 = — X sum of products of columns (6) and (g)
-xol
= o
— 3'75—
B2 = — X sum of products of columns (6) and (/)
Hence
or
_
lol -2-91+3-75)
= -2x2-104 = -421.
y = (o x cos 6) + (o x cos 26) + (4-09 sin ff) + (-42 sin 20)
y = 4-09 sin 0+ -42 sin 26 (the cosine terms being absent).
Analysis by Method (6): The Graphical Interpretation of
(a) (due to Professor Harrison). — To employ this method we must
take at least twice as many ordinates as the largest multiple
HARMONIC ANALYSIS
347
of 0 ', thus if we suppose that the second harmonic is the highest
occurring we might take 6 ordinates as the minimum, although it
would be better if 8 or 10 were taken.
The method can best be illustrated by applying it to an
example.
Example 2. — Resolve the curve ABC [(a) Fig. 130] into its com-
I8O° 24O° 3OO 360*
FIG. 130.
ponent curves (the second being the highest harmonic) ; i. e., find the
values of the constants in the equation
y = AO+AJ cos 6+ A2 cos 20+ A3 cos 30+Bj sin 0+B2 sin 20+ B3 sin 3$.
To arrange that all the ordinates shall be positive, take a base line
DE entirely below the curve. Divide the base into 8 equal divisions
and number the ordinates y0, ylt y2, etc. The angular intervals are thus
^5°, since a full period corresponds to 360°.
348 MATHEMATICS FOR ENGINEERS
Draw a new figure [(&) Fig. 130], the lines OM and ON making 45°
with the principal axes ; number these lines : — o, i, ..... 8, as
shown in the figure. Along line o set off a distance equal to y0, along
line i set off a distance equal to yt and so on. Drop perpendiculars
from the points o, I, etc., on to the principal axes, calling the projec-
tions on these axes h0 (this particular projection being zero), Ax .....
ha, and v0, vl ...... va respectively.
Then to find Aj and Bj : —
As already proved
A! = f (sy cos 0) = "J{y0-y4+ (y!-y3-y&+y7) cos 45°+ (yt-y,) cos 90°}
= MVO-^VI+V^VS-VS+VJ
and similarly Bt = Jj^+Ag— h5— h7+h2— h6}
i. e., the lengths v0 ...... v7, and ht ....... /»7 can be read off
from the figure and then the values of A1 and Bt are calculated as
above.
Jn this example
"0=13-7 *o= o
vv =14 h1= 14
v2 = o h2 = 20-4
v3— ii ha= ii
vt= 7'3 ht= o
v6 = '5 hs= '5
v6= o A6= 1-5
v,= 5-4 h7= 5-4 •*•
andB1 = X38 = 9-5
By the aid of a strip of paper a great amount of this arithmetical
work might be obviated, the procedure being as follows : —
Mark off along the edge of a strip of paper lengths to represent
the various ordinates of the original curve, viz., y0, yv etc., and
number the points so obtained o, i, 2 .... 8 as shown at (c) Fig. 130.
Thus Po == y0, P4 = yt and so on.
We have seen that in order to find the value of Aa it is necessary
to evaluate y cos 6 for the various angles ; i. e., we must find the values
of yv cos 45°, yz cos 90°, ya cos 135° and so on.
Now ys cos 135° = y3x —cos 45° = — ya cos 45°, so that the one line,
viz., that at 45°, serves also for 135° provided that the ordinate is
stepped off in a negative direction. Thus, for example,
yj. cos 6+y3 cos ^6=y1 cos 45°— ys cos 45° = cos 45°(;y1— ys)
and the value of this expression depends upon the difference between
the lengths on the strip Pi and P3, or the distance 3 to i.
Evidently, then, the work is shortened by grouping the ordinates
in pairs to give differences ; thus
y\—y% = Pi— ?3 — i to 3 on the strip
y7— y& — Py— P5 = 5 to 7 on the strip
and so on for other pairs of ordinates.
HARMONIC ANALYSIS 349
Having found these differences, we multiply by 00345°, by setting
these lengths along the line Of in (d) Fig. 130 and then projecting to
the horizontal axis OX ; the resultant of these projections being the
value of 4Aj.
Thus in (d) Fig. 130 : — Make Oo = i to 3 (on the strip) and ab — 5 to 7.
Drop be perpendicular to OX. Then Oc = (y^— y3— y^+y^ cos 45°.
Step off cd = o to 4 (i. e., y^—y^, then measure Od ; this is the value of
4AX, since Od = Oc+cd = (yi—y9—yt+y7) cos 45°+(y0-yt).
A!= 3-43
For the value of Bj the strip must be used according to the following
plan. A line is drawn at 45° [(d) Fig. 130] and distances marked off
along it as follows
Oe = i to 7 on the strip, ef= 3 to 5 on the strip.
{for 46! = (yn+yt) sin o°+(y1+y3-y6-yj sin 45°+(>;2-^6) sin 9<>0}
A perpendicular to OY gives the point g. To Og must be added a
distance = 2 to 6 on the strip, but to avoid extending the diagram this
distance is set off from O giving the length Oh.
Thus Og = 19-5, Oh = 18-2, the sum = 37-7.
Then 4BX = 37-7
and B!=: 9-43.
To find the values of A2 and B2: — The terms containing 26, i. e., 90°,
will now occur and so there will be no lines at 45°.
Ag — |{yo cos °°+yi cos go°+y2 cos 180°+ . . . y1 cos 630°}
Similarly B2 = ^{y!-
Hence set off
Ok = o to 2, i. e., (y9—y%) and kl = 4 to 6, i. e., (yt—y6) along OX
and the resultant is O/ = —-8.
-Hence 4A2=— -8
and A2= —-2.
Set oft
Om =*• i to 3, i. e., (yt—ya) and mn = 5 to 7, i. e., (y^—y-j) along OX
and the resultant is On — — 2-6
Hence 4B2 = — 2-6
and B2 = —-65.
To find the values of A3 and B3 : —
A =-(y° COS °+yi COS I35°+>/2 cos 270°-f-y3 cos 4O5°+>'4 cos 540°
3 §1 +^6cos675°+>/6cos8io0+^7cos9450
-{y9—yt+\y»—yi+yt—y,} cos45°}.
4
350 MATHEMATICS FOR ENGINEERS
Set off Op = o to 4 on the strip, along OX,
and Ob = (y>i— ya— y5+y7) (which has already been done when
finding Aj).
Project b to c on OX ; then Oc = — (ya— y-a-^y^—y-i) cos 45°.
Hence cp = 4A3
but cp = — i -3
and thus A3 = — 33.
To find Bs : —
B _ £j>o sin °°+>'i sin I35°+3/2 sin 270° +y 3 sin 405°+^ sin 540
3 4*- +nsin6750+y6sin8io0+;y7sin9450
= -{yr-yt+fa+yt—yv-yi) sin 45°}-
Set off Oh = 6 to 2 along OY.
O/= (^i+^a— y6— y7) (which has already been done when
finding Bj) .
Project / to g on OY,
then gh = 463
but gh = 1-2
Hence B3 = .3
Also
==
8
(using the trapezoidal rule given on p. 307, Part I),
,• * A 6'75 + I9-8+20-4+i6+7-3+-9+i-9+7-6+6-75
•*•*., A0_ -g—
87-4
^=10-93-
y= 10-93+3-43 cos $+9-43 sin 6— -2 cos 26
~- -65 sin 20 — 33 cos 3$+ -3 sin 3$.
There should be no difficulty in the understanding of this
method if method (a) is first carefully studied. All that this
method (b) adds is the multiplica-
A tion of lengths by the cosines or
sines of angles by regarding the
products as projections on fixed
axes (i. e., if OA = R (Fig. 131)
and the angle AOB = 30°, then
OB = OA cos 30° = R cos 30°, and
OC = OA sin 30° = R sin 30°) .
FIG. 131. The beauty of the method
consists in the use of the strip of
paper for the grouping together of pairs of ordinates which have to
be multiplied by the same quantity.
HARMONIC ANALYSIS 351
In the example just discussed, the angular intervals were taken
as 45°, this choice being made as a matter of great convenience,
since cos 45° = sin 45° and projections may thus be made on either
a horizontal or a vertical axis.
For greater accuracy, more ordinates should be taken, and then
care must be observed as to the axis on which the projections are
made. Thus if the angular intervals were taken as 18°, say, the
lines corresponding to OM, ON and Of in (b] and (d) Fig. 130
would be drawn making angles of 18° with the horizontal axis ;
then for the values of Alt A2, etc., the projections along OX would
be measured, whilst the values of B1, B2, etc., would be determined
from the projections on OY.
Method (c): Analysis by Superposition. — This method is
much used in alternating current work, for the problems of which
it is specially suited. It is not difficult to employ, nor to under-
stand, although the proof of the method is long and is in consequence
not treated here.
In order to present the method in as clear a fashion as possible,
the rules of procedure are here set out in place of a detailed
explanation.
The method is as follows ; the case of a curve containing the
second as the highest harmonic being treated, although the process
can readily be extended if necessary : —
(1) Divide the curve into two 'equal parts and superpose the
second part upon the first, using dividers and paying attention to
the signs. If the resultant curve approximates to a sine curve
there is no need to further subdivide. (This gives terms containing
26, 4$, 60, etc., but if this curve is a sine curve, probably only terms
containing 2$ occur.)
Put in a base line for this new curve (by estimation) ;^ then the
height of this from the original base line = 2A0.
(2) Divide the original curve into three equal parts and super-
pose (first, the second on the first, and then to this result add the
third).
(This gives the terms containing 3$, 60, 'gO, etc.)
The height of the base line of the resulting curve from the
original base line = 3A0. (The two values of A0 may be compared,
and of course they should be alike ; but if not, take the average of
these and draw a new base line distant A0 from the original ; this
line we shall speak of as the true base line.)
(3) Subtract corresponding ordinates of the 20 curve (divided
by^2) and the 36 curve (divided by 3), paying attention to the
352
MATHEMATICS FOR ENGINEERS
signs, from the ordinates of the original curve ; the resultant curve
is approximately a sine curve
symmetrical about the true
base line.
To calculate the values of
the constants, if
+A2sin (20+c2)
A0 is already found.
Select two convenient
values of 6 and work from
the ordinates of the 0 curve
to find Aj and cx; proceed
similarly, using the 26 curve
to find A2 and c2.
Note that in alternating
current work only terms of
the order 6, 36, $0, etc., occur,
so that the curve would need
to be divided into 3, 5, etc.,
equal divisions and the parts
superposed. There is thus
no need to divide into 2, 4,
etc., equal parts; also it is
evident that the value of A0
must be zero.
Example 3. — The curve
ABCD, Fig. 132, gives by its
ordinates the displacement of
a valve actuated by a Gooch
Link Motion.
It is required to find the
constants in the equation
+A2sin
etc.
The original curve is divided
into two equal parts, the second
being placed over the first, with
the result that Curve 2 is
obtained.
The estimated base line for
this is B2 ; the height of B2 above the original base line being -29, i. e.,
HARMONIC ANALYSIS 353
•2Q
the height of the true base line is -- or -145 unit. This base line
can now be put in, and is indicated as the true base line.
By division into 3 and 4 equal parts and superposition the curves
3 and 4 respectively are obtained.
B3, the base line for 3, is at a height of -43 ; this figure divided by
3 gives -143, which agrees well with our former result.
Curve 2 really represents the first harmonic with double amplitude ;
therefore we subtract ordinates of Curve 2 (to half scale, i. e., we use
proportional compasses) from the corresponding ordinates of the
original curve.
Similarly we subtract J of the ordinates of Curve 3 from the
original curve, and since those for Curve 4 are too small to be taken
into account, the net result is Curve i, which represents the funda-
mental, and is a sine curve symmetrical about the true base line.
To find the constants At and c: in the equation
yl = Alsin(6+c1).
When 6 = 0, ;Vi = 2-i75 (measured from the true base line to
Curve i).
At 6 — 90°, y± = o.
c, = 90° or —
2
At 180° yt = —2-135,
t. e., Aj = the mean of 2-175 and 2>I35» *• e-> 2'15>
y = 2-15 sin (0+903)
= 2-15 costf.
To find A2 and c2.
*2Q
The amplitude of Curve 2 is — , ». e., -145.
and since the curve has its maximum ordinate when 6 = 0 we have
again cg = 90, or the curve is a cosine curve.
Hence y^ = -145 cos 20.
Beyond this first harmonic we need not proceed as the amplitudes
of Curves 3 and 4 are exceedingly small.
Hence y = y*+y*
= 2-Ij COS ^+-145 COS 20.
This method of superposition is to be recommended in cases of
A.C. work, as one can so readily tell by its aid which harmonics are
present. If the actual constants in the equation are required it
may be easier to proceed according to method (a) or method (b) .
A A
354
MATHEMATICS FOR ENGINEERS
Exercises 24. — On Harmonic Analysis.
1. Show how to analyse approximately the displacement x of a
point in a mechanism on the assumption that it may be represented
by a limited series of sine and cosine terms, and obtain general expres-
sions for the values of the coefficients in the series
x — 2n (An cos «0+Bn sin n0) + A0
where n = 3 and 6 is the angular displacement of an actuating crank
which revolves uniformly. Apply your results to obtain the values of
the coefficients for the values of x and 6 given in the accompanying
table, where the linear displacement of a point in a mechanism is given
for the corresponding angular displacement of a uniformly revolving
crank.
Angular displacement
of crank in degrees .
o
60
90
120
180
240
300
x (in ins )
I'll
.
•8
1-6
•67
,f
2-Q3
_
2. — A part of a machine has an oscillating motion. The displace-
ments y at times t are as in the table.
t
•02
•04
•06
•08
•i
•12
•M
•16
•18
•2
y
•64
I-I3
i-34
•95
0
— •92
-i-33
— 1-16
-•66
0
Find the constants in the equation
y = A sin (iO7r^+a1) + Bsin (2O7r/+a2).
3. Analyse the curve which results when the following values are
plotted.
x°
0
45
90
135
1 80
225
270
315
360
y
o
21-5
31-25
11-25
0
9
30
26-5
o
4. The values of the primary E.M.F. of a transformer at different
points in the cycle are as follows (6 being written in place of pt for
reasons of simplicity) .
e
o
3°
60
90
120
150
1 80
2IO
240
270
300
330
360
E
-14
886
1293
1400
130?
814
-70
-886
-1293
- 1400
-1307
-814
70
If 6 and E are connected by the equation
E = A sin 0+B sin 3#+C cos 0+D cos
find the values of the constants A, B, C and D.
CHAPTER XII
THE SOLUTION OF SPHERICAL TRIANGLES
THE curvature of the earth's surface is not an appreciable factor
in the calculations following a small survey, and is therefore not
regarded, but when the lengths of the boundaries of the survey are
great, as in the case of a " major triangulation," the effect of the
curvature must be allowed for, if precision is desired. It is there-
fore necessary to use Spherical Trigonometry in place of the more
familiar Plane Trigonometry, and accordingly a very brief chapter
is inserted here, dealing mainly with the solution of spherical
triangles.
Definitions of Terms used. — The earth may be considered
as a sphere of radius 20,890,172 feet, this being the mean radius.
A great circle on a sphere is a circle traced by the intersection
of the sphere by a plane passing through its centre ; if the plane
does not pass through the centre of the sphere, its intersection with
the sphere is called a small circle. Thus all meridians are great
circles, whilst parallels of latitude, except for the equator parallel,
are all small circles.
A straight line on the earth's surface is in reality a portion of a
great circle ; hence a parallel of latitude is not a straight line, or, in
other words, a movement due East or West is not a movement
along a straight line.
A triangle set out on the earth's surface with straight sides is what
is termed a "spherical triangle," its sides being arcs of great circles.
The lengths of these sides might be measured according to the
usual rules, viz., in miles, furlongs, etc., but it is more usual to
measure them by the sizes of the angles subtended by them at the
centre of the sphere. In this connection it is convenient to
remember that an arc of one nautical mile (6076 feet) subtends an
angle of i' at the centre of the earth ; hence a length of 80
nautical miles would be spoken of as a side of 80', i. e., i° 20'.
In Fig. 133 is shown the difference between great and small
circles ; and AB, BC and CA being portions of great circles form a
355
356
MATHEMATICS FOR ENGINEERS
spherical triangle (shown cross hatched). The length BA would be
expressed by the magnitude of the angle BOA.
A spherical triangle ABC is shown in Fig. 134, O being the
centre of the sphere. The arc AB is proportional to the angle
AOB, and therefore, instead of speaking of AB as a length, it is
quite legitimate to represent it by L AOB.
c would thus stand for /.AOB, b for ^.COA, and a for <iCOB.
As regards the angles of the triangle, the angle between CA and
AB is that between the planes AOC and AOB and is, therefore, the
angle between the tangents AD and AE. Spherical triangles
should be regarded as the most general form of plane triangles ;
for if the radius of the sphere becomes infinite the spherical triangle
becomes a plane triangle.
FIG. 133.
Spherical Triangles.
Many rules with which we are familiar in connection with plane
triangles hold also for spherical triangles, as, for example, "Any
two sides of a triangle are greater than the third," or, again, " If
two triangles have two sides and the included angle of the one
respectively equal to two sides and the included angle of the other,
the triangles are equal in ah1 respects " ; " The greater side of every
triangle is subtended by the greater angle."
There is one important difference between the rule for a plane
triangle and a corresponding rule for a spherical triangle: viz.,
whilst the three angles of a plane triangle add up to 180° the sum
of those in a spherical triangle always exceeds 180°, the sum in fact
lying between 180° and 540° ; and the difference between the sum
of the three angles and 180° is known as the " spherical excess."
THE SOLUTION OF SPHERICAL TRIANGLES 357
The magnitude of this can be found from the rule
360° X area of triangle
spherical excess = ^~
2irTz
(27rr2 being the area of the surface of the hemisphere).
This spherical excess is a small quantity for the cases likely to
be considered in connection with surveys.
E. g., consider the case of an equilateral triangle of side 68 miles.
r = 20,900,000 ft. approx. = 3960 miles.
The area of the triangle is about 2000 sq. miles.
Then the spherical excess
360x60x2000
= — — — minutes = -437 minute.
27rX 3960x3960 -"^
A good approximation for the spherical excess of a triangle on
the earth's surface is :
area of spherical triangle in sq. miles
spherical excess (seconds) =
78
Solution of Spherical Triangles. — The most widely used
rule in connection with the solution of plane triangles is the " sine "
rule which states that the sides are proportional to the sines of the
angles opposite. In the case of spherical triangles this becomes
modified and reads — " The sines of the angles are proportional to the
sines of the sides opposite.
Therefore, adopting the notation of Fig. 134,
sin a sin b sin c
sin A ~ sin B sin C
it being remembered that sin a is really sin L BOC, etc.
Other rules are
. A /sin (s— b) sin (s— c)
sin — = A/ - v . ' —i- .... (2)
2 >
.
2 > Sln f) sln c
A /sin s sin (s— a)
cos — = \/ — : — =—± — .... (3)
2 v sin & sin c
A /sin (s— 6) sin (s— c)
tan— =v — ^ : , v . . . . . (4)
2 v sin s sin (s— a)
T> /-»
and corresponding forms for — and - , obtained by writing
2 2
the letters one on in the proper sequence, a b c a.
s in these formulas = — - and is, therefore, an angle
(in plane trigonometry, s = - — , but is a length).
358 MATHEMATICS FOR ENGINEERS
It is of interest to compare these with the corresponding rules
in connection with plane triangles, which are
• A /(s—b)(s—c)
sin - = \/ '- '-
2 v be
A A
'*=V-
's(s —
COS
be
8(9 -a)
It will be seen that, as in the previous case, sides occurring in
the formulae of plane trigonometry are replaced by their sines in
the corresponding formulae of spherical trigonometry.
Other rules are : —
cos A+cos B cos C
cos a = . . (5)
sin B sin C
cos a — cos b cos c
cos A — : — =—. — (6)
sin b sin c
cot A sin B = cot a sin c — cos B cos c . . . (7)
(a-V)
cos-
tan~lT~ ~7^R>\COt2 (8)
COS - — J
\ 2 /
(a-b)
. sin —
A— B 2 C
tan — — = - — -r- cot —.....-.. (q)
2 . fa-\-o\ 2
sin -
V 2 /
Solution of Right Angled Spherical Triangles. — In the
case of a right angled spherical triangle these rules can be put into
somewhat simpler forms.
Assume that the triangle is right angled at C.
€ = 90°, .'. cosC = o, and sinC = i.
„ cose— cos a cos b
From (6) cos C =
sin a sin b
but cos C = o,
cos c— cos a cos b = o,
i.e., cos c = cos a cos b (10)
THE SOLUTION OF SPHERICAL TRIANGLES 359
cos a — cos b cos c
Also cos A = - — . . . — from (6)
sin 0 sin c
cos c
= f — COS 0 COS C
COS0 ,
— from (10)
sin 0 sin c
cos c i cos 0 cos c
— \s _^_, w
sin c sin 0 cos 0 sin 0 sin c
if i cos 01
~ tan c\sin b cos b sin 0 J
:— cos2 &1 cot ex sin2 0
= cote] T
Isi
sin 0 cos 0J sin 0 cos b
sino
= cot c X ,
COS0
= cot c X tan 0
or tan ox tan (90— c)
i.e., cos A = tan b tan (90 — c)l , >
also cos B = tan a tan (90— c) J
COStf — COS 0 COS C
Again cos A = - — ; — ^—. —
sin & sin c
COS2C
cos a —
cos a , . .
from (10)
sin 0 sin c
cos2 « — cos2 c sin2 c — sin2 a
= —•-»-• — or — ~ir- — • • (12)
cos a sin 0 sin c cos a sin 0 sin c
. . sin a
And from (i) sin A = -, — (13)
sine
In plane trigonometry sin A = - .
Napier's Rules of Circular Parts. — The equations (10),
(n) and (13) and their modifications may be easily remembered by
Napier's two rules of circular parts, which may almost be regarded
as a mnemonic.
For the application of these two rules the five parts of the
spherical triangle, other than the right angle at C, are 'regarded as
a, b, (go — A), (90— c), and (90— B) respectively, the complements of
A, c and B being taken instead of the values A, c and B in order
that the two rules may embrace all the cases.
360 MATHEMATICS FOR ENGINEERS
These five parts are written in the five sectors of a circle in the
order in which they occur in a triangle : thus in Fig. 135, com-
mencing from the side a and making the circuit of the triangle in
the direction indicated, the parts in turn are a b A (for which we
write 90— A), c (for which is written 90— c) and B (for which is
written 90— B). These parts are set out as shown in Fig. 136.
Then Napier's rules state : —
Sine of the middle part = product of tangents of adjacent parts.
Sine of the middle part = product of cosines of opposite parts.
The terms middle, adjacent and opposite have reference to the
mutual position of the parts in Fig. 136. Thus if b is selected as
the middle part, the adjacent parts are those in immediate contact
FIG. 135. FIG. 136.
with b, viz., a and (90— A), whilst (90— c) and (90—6) are the
opposite parts.
Hence sin b = tan a x tan (90— A) = tan a cot A
sin b = cos (90—6) X cos (90— c) = sin B sin c
. „ sin b .„, ,. , .
or smB = -— . (Cf. equation (13), p. 359.)
olll C/
Again if (90—6) is selected as the middle part, the adjacent
parts are (go—c) and«, and the opposite parts are (90— A) and b.
Hence sin (90—6) = tan (90— c) tana
or cos B = tan (90—0) tan a (cf . equation (n), p. 359),
and sin (90— B) = cos (90— A) cos b
or cos B = sin A cos b.
These rules, being composed of products and quotients only,
lend themselves well to logarithmic computation.
The Ambiguous Case in the Solution of Spherical
Triangles. — In the solution of a plane triangle, if two sides and the
angle opposite the shorter of these is given, there is the possibility
THE SOLUTION OF SPHERICAL TRIANGLES 361
of two solutions of the problem ; the best test for which, as pointed
out in Chap. VI, Part I, being the drawing to scale.
A similar difficulty occurs in the solution of spherical triangles,
when two sides and the angle opposite one of them is given.
E. g., let a b and B be the given parts.
Then from equation (i), p. 357.
sin A sin B
sin a sin b
sin a sin B
or sin A = : — -, — .
sino
Now sin A = sin(i8o— A) and thus the right-hand side of this
last equation may be the value of either a particular angle or its
supplement.
Without going into the proof it may be stated that there will
be one solution only if the side opposite the given angle has a
value between the other given side and its supplement. Thus in
the case in which a b and B are given, there will be one solution
only if b lies between a and (180— a}.
If b is not between a and (180 — a], then the test must be
applied that the greater angle must be opposite the greater side :
thus for the case of a b and B given, if a > b then A must be > B.
The possible cases may be best illustrated by numerical examples,
a b and B being regarded as the given parts throughout.
(a) Given a = 144° 40', b = 87° 37', B = n°g' to find A.
Using equation (i) of p. 357
A _ sin a sin B _ sin 144° 40' X sin 11° 9'
Sin A : f ; = -.
sm b sin 87 37
and log sin A = log sin 144° 4o'-flog sin 11° 9'— log sin 87° 37'
= i • 7622 + 1 • 2864—1" • 9996
= I • 0490
so that it is possible that A = either 6° 26' or 173° 34'.
Now a > b and therefore A must be > B, and this condition is
only satisfied if A = 173° 34', since 6° 26' is not > n°9'.
It will be noted that the case chosen is that in which b, viz.,
87° 37', lies between a, i. e., 144° 40', and (180— a), i. e., 35° 20', and
therefore only one solution is expected.
. (b) b = 44° 35', a = 55° 10' and B = 38° 46'.
Here b does not lie between a and (180— a), so that two solutions
are possible.
362 MATHEMATICS FOR ENGINEERS
As before
log sin A = log sin a + log sin B — log sin b
= log sin 55° io'+ log sin 38° 46'— log sin 44° 35'
= I • 9142+1 • 7966—1 • 8463 = I • 8645
— log sin 47° 3'
so that possible values of A are 47° 3' and 132° 57' and we must
test each of these values.
Now a > b, and hence A must be > B ; but 47° 3' and 132° 57' are
both > 38° 46', so that we have two triangles satisfying the con-
ditions, and for complete solution the two values of A, C and c
must be determined.
Example i. — In a spherical triangle ABC, having given 0 = 30°,
6 = 40°, C = 70°, find A and B.
Given also that
Lsin 5° =8-9402960 L tan 12° 14' 38" = 9-3364779
L sin 35° = 9'75859i3 L tan 60° 4' 3" = 10-2397529
L cos 5° — 9-9983442
L cos 35° = 9-9133645
In this case two sides and the included angle are given ; we there-
fore use equations (8) and (9) .
fa— b\
A+B DSV 2 / ,C , ,„.
tan — — = — — cot - . . from (8)
2 (a+b\ 2
cos f — — j
=-5_:cot35. {*»<^f }
cos 35° \
cos 5° cos 35° _ cos
cos 35° sin 35 °~ sin 35°-
Taking logs of both sides
A+B
Ltan — — = L cos 5°— Lsin 35°+io
A+B
(or, alternatively, log tan — — = log cos 5°— log sin 35°)
Ltan
A+B I9-9983442
— — =
= L tan 60° 4' 3*
A+B
-^- = 6o°4'3"
A+B = 120° 8' 6" ....... (a)
THE SOLUTION OF SPHERICAL TRIANGLES 363
From equation (9)
• fa-b\
. -_. sin I 1 „
A— B \ 2 / , C
tan — — —-j- cot
2 . fa+b\ 2
sin -
V 2 /
A— B_ sin 5° 00535°
tan
sin 35" sin 35°
B— A _ sin 5°x cos 35°
sin2 35°
taking logs throughout.
B— A
L tan — — = Lsin 5°-fLcos35°— 2Lsin35°+io
18-9402960
9-9I33645
= 28-8536605
19-5171826
=- Ltan 12° 14' 38*
B-A=24°29'i6* ...... . (6)
By adding (a) and (b) 2B = 144° 37' 22*
B = 72° i8'4i*
and A = 120° 8' 6"— 72° 18' 41" = 47° 49'25*.
[Note that A+B+C = 47° 49' 25"+72° i8'4i*+7o°
= 190° 8' 6'
so that the spherical excess = 10° 8' 6".]
Example 2. — Solve the spherical triangle ABC, having given
c = gi°i8', a — 72° 27', and 0 = 90°.
In this case the triangle is right angled, and therefore rules (10) to
(13) may be used.
To find A :—
. . sin a
From equation (13), p. 359, sin A— -: --
sin o
L sin A = L sin a— L sin c-\- 10
= Lsin 72° 27'— Lsin 91° i8'+io
19-97930
9-97941
= L sin 72° 29' 45*.
A = 72° 29' 45'.
364 MATHEMATICS FOR ENGINEERS
To find b : —
From equation (10), p. 358,
cos c — cos a cos b
cos c
whence cos 6 = -
cos a
cosoi°i8/ —cos 88° 42'
* cos b — • _ — _ — _ -- - _
cos 72° 27' cos 72° 27'
cos 88° 42'
or cos (180— b) = — cos 6 = - — 5-^— >•
cos 72° 27'
Hence we shall work to find the supplement of 6.
Taking logs *
log cos (180— 6) — log cos 88° 42'— log cos 72° 27'
= 1147934
2-87644
= log cos 85° 41' 7*.
180—6 = 85° 41' 7"
* It is rather easier to work in terms of the logs in preference
to the logarithmic ratios. One must remember, however, that the
L sine A = log sin A+io, so that if a L sin A reading is 9-97941, then the
reading for log sin A would be 1-97941. If the logarithmic ratios are
used the addition of the 10 must not be overlooked.
To find B :—
From equation (n), p. 359,
cos B = tan a tan (90°— c)
i. e,t cos (180°— B) =tanatan (c— 90°) = tan 72° 27' x tan i° 18'.
Iogcos(i8o— B) — log tan 72° 2 7'+ log tan i ° 18'
•4990
= 2-3559Q
2-85586
-log cos 85° 53' 6".
180—6 = 85° 53' 6"
B = 94°6'54*.
Hence, grouping our results,
a = 72° 27' A = 72° 29' 45'
6-94°i8'53* B = 94°6'54*
C = 90°
Example 3. — At a point A, in latitude 50° N., a straight line is
ranged out which runs due E. at A. This straight line is prolonged for
THE SOLUTION OF SPHERICAL TRIANGLES 365
60 nautical miles to B. Find the latitude of B, and if it be desired to
travel due N. from B so as to meet the 50° parallel again at C, find the
angle ABC at which we must set out and also the distance BC.
In Fig. 137 let A be the point on latitude 50° N. and ABD be a
great circle passing through A : thus AB
is a straight line running due E. from A.
Let NB be the meridian through B, and NA
that through A.
The sides NA, AB and BN are straight
lines, because they are parts of great circles
and therefore they together form a spheri-
cal triangle.
In this triangle we know -the side AB
(its value being 60', for i nautical mile
subtends an angle of i' at the centre) ;
the angle at A (90°) ; and the side NA
(90°— latitude, i.e., 40°).
Thus two sides and the included angle are given and we require to
solve the triangle ; hence we use rule (10), p. 358,
from which cos NB — cos NA cos AB
= cos 40° cos 60'
or logcosNB = logcoS4O°-f log cos i°
= 1-88425 +1-99993
= 1-88418
i.e., NB = 4o°o'38'1'
or the latitude of B is
Q0°— 40° o' 38* = 49° 59' 22*.
Now C is at the same latitude as A, so that BC is 38", corresponding
to f- nautical miles ; i. e., BC — -633 nautical mile.
60
To find the angle ABC, we use rule (13).. p. 359.
. __ sin NA sin 40°
SI* L. ABC = ,-== = • . ^-f—Q,
sin NB sin 40 o 38*
log sin L. ABC — log sin 40°— log sin 40° o' 38*
— i -90807 — I -90817
= I -99990
whence L ABC = 88° 45'.
For the surveyor, spherical trigonometry has an important
application in questions relating to spherical astronomy. Thus in
the determination of the latitude of a place by observation to a
star, the calculations necessary involve the solution of a spherical
triangle. This triangle is indicated in Fig. 139, the sides AB, BC
366
MATHEMATICS FOR ENGINEERS
and CA representing the co-latitude of the place, i. e., (90°— latitude),
(90° — declination) and (90°— altitude) respectively ; whilst the
angles A, B and C measure respectively the azimuth, the hour
angle and the parallactic angle.
The terms just mentioned are denned as follows:—
Fig. 138 represents a section of the celestial sphere at the
meridian through the point of observation O. RDT is the celestial
equator, CEX is the horizon, Z is the zenith of the point of
observation, i. e., the point on the celestial sphere directly above O,
and S marks the position of the heavenly body to which observa-
FIG. 138. — Determination of Latitude.
tions are made. Also PSD, ZSC, RDT and CEX are portions of
great circles.
The altitude of a heavenly body is the arc of a great circle
passing through the zenith of the point of observation and the
heavenly body ; the arc being that intercepted between the body
and the horizon. We may thus compare the altitude in astro-
nomy with the angle of elevation in surveying. Referring to
Fig. 138, ZSC is the great circle passing through Z and S, and SC
is the altitude. ZS, which is the complement of SC, is called the
zenith distance.
THE SOLUTION OF SPHERICAL TRIANGLES 367
The azimuth of a heavenly body is the angle between the
meridian plane through the point of observation and the vertical
plane passing through the body. It can be compared with the
'bearing" of plane trigonometry. In Fig. 138, the angle PZS is
the azimuth of S.
The hour angle of a heavenly body is the angle at the pole,
between the meridian plane through the point of observation and
the great circle through the pole and the body.
Thus, in the figure, P is the pole, and PSD is the great circle
passing through P and S ; this being known as the " declination
circle." Then ^_ZPS = the hour angle of S, and it is usually
expressed in terms of time rather than in degrees, etc.
The declination of a heavenly body is the arc of the declination
FIG. 139.
circle intercepted between the celestial equator and the heavenly
body : thus DS is the declination of S.
The method of calculation can be best explained by working
through a numerical example ; and in order to ensure a clear
conception of the problem, it is treated both graphically and
analytically.
Example 4. — At a certain time at a place in latitude 52° 13' N. the
altitude of the Sun was found to be 48° 19' and its declination was
15° 44' N. Determine the azimuth.
As explained before, a spherical triangle can be constructed with
sides as follows: a — 90— declination = 74° 16', b = 90— altitude —
41° 41', and c = co-latitude — 37° 47'. Then the angle A is the
a?imuth (Fig. 139).
368 MATHEMATICS FOR ENGINEERS
Graphic construction. — With any convenient radius OD describe an
arc of a circle DABF. Draw OA, OB and OF, making the angle
DOA = 6 = 4i°4i', LAOB = c = 37° 47', and L BOF = a == 74° 16'.
Draw DCE at right angles to OA, and FGC at right angles to OB,
intersecting at the point C. Note that C lies outside the triangle AOB.
With centre E and radius ED construct the arc of a circle DH : draw
CH perpendicular to DE to meet this arc at H and join EH. Then the
angle REH is the value of the required angle A, and is found to be in
the neighbourhood of 141°. [If C had fallen the other side of OA, the
angle CEH would have been measured.]
The actual spherical triangle ABC is formed by the circular arc
BA and the elliptical arcs AC and BC.
Proof of the construction — The side b is such that it subtends an
angle of 41° 41' at the centre of the sphere. Thus DA measures the
actual length of b, but does not represent it in its true position. In like
manner BF gives the length of a, but again does not give its position
on the sphere.
Let the circular sector OAD be rotated about OA as axis, and the
sector OBF about OB as axis, and let the rotation of both be continued
until they have a common radius OC, i. e., OC is the intersection of the
two revolving planes. Then evidently C is the third angular point of
the spherical triangle ABC, since the given conditions concerning the
lengths of the sides are satisfied by its position.
We observe that in this case the rotation of OBF has to be continued
beyond OA, from which fact we gather that the angle at A must be
obtuse. The line OA is in the plane of the paper, and taking a section
along DE and turning this down to the plane of the paper, we observe
that the actual height of C above the paper is CH. Thus EH is a line
on the plane OAC, also ER is a line in the plane AOB, both lines being
perpendicular to the line of intersection, and the angle REH therefore
measures the inclination of the plane AOC to the plane AOB, this angle
being by definition the angle A of the spherical triangle ABC.
By calculation.— Here we have the three sides given, and we wish to
find an angle which may be done by use of equation (4), p. 357, viz.,
A /sin (s—b) sin (s— c)
tan = A / - : - : - - - r --
2 N sin s sin (s—a)
Now , = <*+*>+<> _ 74° i6'+4i° 4i'+37° 47' _ 153° 44' ^ 6o ,
222
so that s— a— 76° 52'— 74° 16' = 2° 36'
s-b = 76°52'-4i04i' = 35° ii'
Hence tan A = /^
2 v sin 76° 52' X sm 2° 36
THE SOLUTION OF SPHERICAL TRIANGLES 369
A i r(logsin35°Ti/+logsin39°5') ~1
and log tan - = - [_ _ (log sin ?6o 52/+iog sin 2» 3g/)J
i'76°57\
1-79965 J _ f 2-65670
I- 56022 /
= £x -91503 = H5752
thus — = 70° 46' 30*
and A = 141° 33'.
Exercises 25. — On the Solution of Spherical Triangles.
(4-figufe log tables only have been used in the solution of these
problems.)
In Nos. i to 6, solve the spherical triangle ABC, when
1. « =^50° 0 = 90° & = 32°i7'.
2. 0 = 90° a = 45°43' A = 6i°i5'.
3. a = 72° 14' & — 43° 47' c = 29° 33'. Find also the value of B
by the graphic method explained on p. 368.
4. Z> = 52°5' a = 58° 25' C = 64°.
5. b = 27° 13' c = 5i°i8/ 6 = 85° 9' and the spherical excess is
2° 14'.
6. c = 79°49' b = 28° 5' B=i5°i8'
7. If the sun's altitude is 17° 58', its declination is 28°i6'N., and
its azimuth is N. 65° 43' W., find the latitude of the place of observation.
8. The spherical excess of a triangle on the earth's surface is i° 15':
taking the earth as a sphere of radius 3,960 miles, find the area of the
triangle in square miles.
9. Given that the azimuth of the sun is 10°, and its zenith distance
is 24° 50' when its declination is 22° 15', find the latitude of the place
and also the hour angle.
B B
MATHEMATICAL PROBABILITY AND THEOREM OF
LEAST SQUARES
WHEN extremely accurate results are desired, these results
being derived from a series of observations, the possibility of error
in each or all of the observations must be considered. The correct
result, or what is termed " the most probable result," is usually
found by combining the mean of the observations with " the pro-
bable error of the mean." The work that is to follow is concerned
primarily with the establishment of a rule enabling us to find this
probable error ; and as a preliminary investigation, a few simple
rules of probability will be discussed.
Supposing that an event is likely to happen 5 times and to
fail 7 times, then the probability that it will happen on any
specified occasion is r\, whilst its probability of failing is r7^,
because, considered over a great range, it only happens 5 times
out of 12. It is important to note the significance of the phrase
*' considered over a great range " ; we could not say with truth that
the event was bound to happen 5 times out of the first 12,
10 times out of the first 24, and so on ; it might be doubtful
whether it would happen 50 times out of 120. If, however, say,
12,000 opportunities offered, it would be fairly correct to say that
the happenings would be 5,000 and the failures 7,000, for when a
large number of occasions were considered, all " freaks " would be
eliminated.
To take another illustration : — the probability that a man will
score 90 per cent, of the full score or over on a target is /y
indicates that he is rather more likely to score 90 per cent, than
not (in the proportion 6 to 5) if he fires a great number of shots.
In general terms, if an event may happen in a ways and fail in
b ways, and all these are equally likely to occur, then the pro-
bability of its happening is -j^-, and of its failing — =- ; and if
37°
MATHEMATICAL PROBABILITY 371
a = b, then it is as likely to happen as not, i. e., its probability of
either happening or failing is \.
„ Probability of happening _ a a+b _ a
Probability of failing ~ a+b b ~ b'
i. e., the odds are a to b for the event, or b to a against it, the
first form being used if a > b and vice versa.
E.g., if the odds are 10 to i against an event, the probability of
its happening = — — = — ; or it will probably happen once only
out of eleven attempts.
Exclusive Events. — Let us now consider the case of two
exclusive events, viz., the case in which the happenings do not
concur.
/»
Suppose the probability of the happening of the first event =-
/
and the probability of the happening of the second event = -T-.
Then for purposes of comparison each of these fractions may be
expressed with the same denominator : if this common denominator
is c, write the fractions as — and — respectively.
c c
Now out of c equally likely ways the first event may happen in
a± ways and the second in «2 ways, and since the two events are
exclusive, i. e., the happenings of the one do not coincide with the
happenings of the other, the two events together may happen in
«!-}-«, ways.
Hence the probability that one or the other will happen is
— -. which may be written in the form — -f- — , i. e., as the sum
c c c
of the separate probabilities.
E. g., suppose that one event happens once out of 8 times, and
a second event happens three times out of 17, and that there is no
possibility of the two events happening together ; then, the
common denominator of 8 and 17 being 136, the first event happens
17 times out of 136 and the second event happens 24 times out of
136, and hence, either the one or the other happens 41 times out cf
each 136.
Probability of the Happening together of Two Inde-
pendent Events. — Suppose that one event is likely to happen
once out of every 6 times, whilst another is likely to happen twice
out of every 17 times ; then the probability that the two will
happen together must be smaller than the probability of the
372 MATHEMATICS FOR ENGINEERS
happening of either — in fact, it must be the product of the separate
probabilities ; i. e., the probability of the two events happening
122 I
together = ;rX — = or — : or out of every 10.200 times the
6 17 102 51
first will probably happen 1,700 times, the second will probably
happen 1,200 times, whilst the two would happen together 200
times only.
Probability of Error. — Bearing in mind these fundamental
theorems, we can proceed to a study of the question of probability
of error ; with particular reference to its application in precision
surveying.
It will be admitted that, for any well made series of observa-
tions, the following assumptions may be regarded as reasonable : —
(1) That small errors are more likely to occur frequently than
large errors, and hence extremely large errors never occur.
(2) That positive and negative errors are equally likely, i. e., we
are as likely to give a result that is -ooi too high as one that is
•ooi too low.
Hence the probability of the occurrence of an error of a given
magnitude, which is denoted by
the number of errors of that magnitude
total number of errors
depends in some way upon the magnitude of that error. Our first
idea, therefore, might be that the probability of the occurrence
of an error of magnitude x could be expressed as f(x), i. e., as some
linear function of x. It will be seen, however, that this is not in
accordance with assumption (2) ; for assumption (2) demands that
if a curve be plotted, the ordinates showing probabilities and the
abscissae indicating errors, it must be symmetrical about the y axis.
The function must therefore be of an even power of x, and taking
the simplest power we say that the probability of occurrence of an
error of magnitude x =y = f(x2).
Now, from assumption (i) we note that the coefficient of xz
must be negative, because y must decrease as x increases.
The probability of an error of magnitude x being included in the
range x to x +S# must thus depend on xz, and also on the range Bx ;
hence it would be reasonable to say that it =f(xz}8x, because the
greater the range the more is the chance of happening increased.
Therefore, the probability that an error of magnitude x falls
MATHEMATICAL PROBABILITY 373
between any assigned limits, —a and -\-a, must be the sum of the
probability f(xz) 8x extended over the range — a to -\-a,
i.e., P = ^ f(x*)dx
this being the probability that the error does not exceed a.
Hence the probability that the error may have. any value
whatever (i. e., the probability is i) must be expressed by
for the range is unlimited, so that
/•+«
I f(x2)dx must = i.
.' -oo
It has been proved by Lord Kelvin (see his " Natural Philo-
sophy ") that f(x2) must be such that
and since e?* x 0* =
and e1** X e*v* =
this condition will be satisfied if
_«2
or Ae~*2
the minus sign being inserted in accordance with assumption i on
p. 372 ; and the coefficient k being written as ^ f°r the reason that
ft
is explained later.
A value can now be found for the constant A.
r+oo
r+oo
It is known that I f(x2)dx = i,
J -00
/• + » _Z2
hence A / e &dx — i.
Now it has already been proved (see p. 163) that
r -z* ™
I Oe ' *=''~z~
/•»_?: " h^/ir
and I e h*dx =
Jo 2
>r2 r2
/•OO _ Z^ ,-flO _ *"
also I e h'2dx = 2 / e A2rfA; =
y -oo J o
374 MATHEMATICS FOR ENGINEERS
hence AxWir = I,
or
A =
Thus
y =/(*•) =
the law being known as the Normal Error Law.
The curve representing this equation is called the probability
curve and also Gauss's Error curve. Two such curves are plotted
in Fig. 140, to show the effect of the variation of the parameter h.
In the one case h = -2, and for the second curve h = -5 ; and it
will be noticed on comparing the curves that for the smaller value
of h the probability of the occurrence of small errors is greater,
i. e., the set of observations for which h = -2 would be more nearly
correct than that for which h = -5.
It will be seen that the curve is in agreement with the axioms
stated on p. 372 ; for the probability of error is greatest when the
error is least, the probability of a large error is very small, and
there is as much likelihood of an error of + -2, say, as of —-2.
The probability that the error does not exceed -i is given by
the area ABCD in the one case, and ABEF in the other.
Theorem of Least Squares. — If a number of observations are
made upon a quantity, and the errors in each of these noted, i. e., as
nearly as can be estimated ; then from a knowledge of these errors
it is possible to find the most probable or likely value of the
quantity.
Let n observations be taken and let the errors be x^ xz • • • Xn '.
also suppose that all the measurements are equally good, i. e., the
"fineness" of reading is the same throughout; h in the formulae
above being a measure of the fineness.
The probability of the error x± being within a certain range 8x
MATHEMATICAL PROBABILITY 375
will be the probability of an error of magnitude xl multiplied by
the range «*, i.e.,
i .
and for error x2 P2 = 8xX — -=.e~ *a and so on.
Now #! x2 etc. are quite independent, so that the probability of
all the errors falling within the range &x will be the product of the
separate probabilities, i.e.,
P = P1XP8X . . .PB
&x -^ Sx -**
x — =e »2 X . .
hVi
We have thus obtained an expression which gives us the prob-
ability of all the errors falling within a certain range. We might
say that this range was -i, for instance, or -05. Evidently if all
the errors were kept within the range —-05 to +-05 the calculated
result would be a nearer approximation to the truth than if the
range were double the amount stated.
Our object then is to find when the probability of a small error
(8x may be reduced as we please) is greatest, *'. e., to find when P is
a maximum.
_ .. -!(*-) K
Now P = K e h& — -;
and the smaller the denominator is made, the larger will P become.
But the only variable in the denominator is 2#2, and hence, in order
that P may have its maximum value, 2#2 must be the least
possible. Hence the most probable value of the quantity to be
determined is that which makes the sum of the squares of the
errors the least.
The fact can now be established that the arithmetic mean of
the observed values is the most probable value of the quantity.
376 MATHEMATICS FOR ENGINEERS
Thus, if n observations are made,
let «! «2 #3 . . . On be the respective observations
a the A.M. of these values
a the value most probably correct
then («! — a) (a2—a) etc. are known as residual errors.
Now the probability of making this system of errors
T> A ~ ro
or P = Ae h~
- A -
To differentiate P with regard to a, put u =
du
then i_ = o+2tfw
fl#
Thus P = A*~^
rfP dP du
_ __ _ \/ _
da du da
A ---
= — F5
h2
dP
and -3-=- =o if 2an— zSa, = o
w
or if « = -za-,
n
but -2#i = a = A.M. of the observations
n
and hence <z = a
or the most probable value is the A.M. of the observations.
Again, if x is the error of the A.M. and xl xz x3 etc. are the
respective errors of the observations,
By squaring
X = -
ft
= l2(2V)+i<
M2V I / l nz\
MATHEMATICAL PROBABILITY 377
then, since it is assumed that all the observations are equally good,
and that positive and negative errors are equally likely to occur,
v 2 — -y 2 — A* 2 — — ,,- QTtin ^? ^ v — {\
»vj — •^'2 — 3 — * " • — r^ ciuA-i ^H-vi-vo "~~— vj
for all the errors are small and their products, two at a time, are
still smaller.
i u2
Also x2 = — («u2) = —
«* «
or # = ~=
Vn
. . probable error of a single observation
or the probable error of the A.M. = — ,
Vn
and thus, other things being equal, the possibility of a large error
in the final result is greatly reduced by taking a great number of
observations. Also in a set of well made observations, if a sufficient
number are made, the arithmetic mean cannot differ from any of
the observations to any very great extent, and accordingly the
residual errors and .the actual errors are very nearly alike.
We are now in a position to summarise the results of the
investigation so far as we have pursued it ; thus
(a) The arithmetic mean of the series of observations, which are
supposed to have been made with equal care, is the most probable
value of the quantity.
(b) The sum of the squares of the residual errors must be the
least.
(c) The probable error of the A.M. is equal to the probable
error of a single observation divided by the square root of the
number of observations.
Example i. — Seven observations of a certain quantity, all made with
equal care, were 12, n, 14, 12, 11-2, 11-7, and 12-1.
Find the most probable value of the quantity.
The most probable value = A.M. = — = 12,
7 —
and it can readily be shown, by actual calculation, that this value
makes the sum of the squares of the residual errors the least.
The residual errors are
(12 — 12), (11 — 12), (14—12), (12 — 12), (11-2 — 12), (11-7—12)
and (12-1 — 12) or o, — i, 2, o, —-8, —-3, -i
and 2 squares of residual errors — 0+1+4+0+ -64+ -09+ -01
= 574-
378 MATHEMATICS FOR ENGINEERS
To test whether this is the least, let us suppose that the most
probable value is 11-5; then the residual errors are: -5, —-5, 2-5, -5,
— •3, -2 and -6 respectively.
2 squares = -25 + -25+6-254- -25+ -09+ -04+ -36 = 7-49.
Similarly, if we assume, say, 12-2 as the most probable value,
Z (residual error)2 = (.2)2+(i-2)?+(i-8)2+(-2)2+(i)2+(-5)2+(-i)2
= -04+1-44+3-24+ -04+1+ -25+ -oi
= 6-02
both of which totals exceed 5-74.
To find the Probable Error of the Arithmetic Mean.—
Let r = the probable error of any one of the observations ; then if
this is an "average" error, i. e., if errors greater are as likely to
occur as errors smaller, the probability that the error is less than
r is \.
Now, the probability that an error lies within the range — r to -\-r
T f+r -I2 2 [' -r-
is — =1 e h*dr = — -=.1 e h2dr
T f+r -I 2 ['
— =1 e h*dr = — -=.1 e
hV-n-J -r flVirJ 0
- /r
/for dr=hd( r ) and the limits are now those for T and not those for r\
\ \hj h I
There must be some connection between the amount of error
and the fineness of measurement, i. e., between r and h, and this we
must now find.
If X = £
h
A/V' o x«x V ir-i o
and we see from the above statement that the value of this
integral is to be £.
9 o
XT , X * , X3 .
N o w e — i — i- jf -i u — i—
2 6
and thus e~X2 — i— X2-| ^-+ .
2 o
and if X2 is small we may perform the integration by way of
expansion in series : if X2 is not small the value of the integral
would be read from probability tables which give the values of the
2 /-x _X2
integral — ^ I e ^X : these tables being given in the Transactions
VWo
of the Royal Society of Edinburgh, Vol. xxxix. For the present
MATHEMATICAL PROBABILITY
379
application of the integral, however, X is a small quantity, and a
sufficiently correct result is obtained by expanding in a series and
calculating from a few terms in this series.
Thus
L _T
o fh -X2 -2 / rh --' —~
-41 « dX = -2( (i
VTT\J 0
FIG. 141.
Hence
and this equation may be written
v5 = /r_j:L . '
¥
or if for j- we again write X
X3 X5 X7
•44SI = X 1 h terms which are very small.
3 10 42
By selecting values of X and plotting, the solution of this
equation is found, the final plotting being represented in Fig. 141,
where it is seen that the solution is X = -4769.
38o MATHEMATICS FOR ENGINEERS
Y
Thus T = '4769 or r = -4769^.
Y
[If solved to a greater degree of accuracy, the value of r is
found to be -47696^, and this figure will be used in the work that
follows.]
Again, if n equally good measurements have been made, each
will have what is termed a weight of unity, i. e., none is better or
worse than any other, and when working towards the result to
be deduced from the measurements, equal consideration must be
paid to each measurement ; also the A.M. is said to have a weight
of n since on the average n observations of equal weight must be
made to give a result as true as the A.M.
r
Knowing that rm = ~/^
Vn
where rm = probable error of the A.M.
and r = probable error of any observation
weight of A.M. n
and also — r-^-r — ?- *r- —r- — = -
weight of one observation i
, . , .. v» «
which we can write as — = -
w i
we can link up wm and w with rm and r,
for ^!-^i
r* n wm
or the weight varies inversely as the square of the probable error.
Thus the determination of the probable error, whilst a useful
guide to the accuracy of the one set of observations, is more use-
ful in fixing the relative weights that must be given to different
sets of observations.
Thus, if three sets of observations have been made on a certain
length with the results that the probable errors of the A.M. are
•45, -29, and -51 respectively ; then the weights to be given to
these sets are - — ^ -. — -^ - — r^ respectively
(.45)2 (.29)2 (-5i)2
or -494 1-19 -384.
Then in assessing for the final result, by far the most reliance
would be placed on the second set of observations, less on the first,
and least on the third set ; this fact being well illustrated by
MATHEMATICAL PROBABILITY 381
Fig. 142, the resultant weight being nearer to the weight 1-19 than
to either of the other weights.
1R.2-068
•494 1-19 394
FIG. 142.
To return to the object of this paragraph : — •
If #! #2 *s • • • are the actual errors of observation, then the
probability that each falls within the small range 8* is
i l^2 i -(^Y
P! = — ^j V*/ 8* P2=— -7=e ^h) (&) etc.,
hVir hVir
and the probability that they all fall within this range at the same
time will be less than either of the separate probabilities ; it will
actually be the product of these.
Thus P = P1xP2X ......
i (
~
hVir hV-jr
-(*,«+*,«+
h
We wish to find for what value of h P has its greatest value
hence differentiate P with respect to h.
P xe-- where K =
= UXV
= and
hn dh
v = g-A2<2a:i2) or if w = TtCZx-,2) v =
h?
and thus ~ = (2^2) X -2h - 3
ah
dv dv dw
Also — = -- x -3-
an aw dh
_de-" dw_ ^
: dwXdh~
382 MATHEMATICS FOR ENGINEERS
Then
dP du . dv
~rr= v-r; +U-T;
dh dh dh
=o if
22V
\/^
or h== 1-414
Now it has already been proved that r = -47696/1
so that r = -47696 Vs
/(W)
- -6745V ^
Also we have previously stated that the sum of the squares
of the actual errors differs very little from the sum of the squares
of residual errors ; this being true if a great number of observations
are taken. The difference in the two sums may be expressed rather
more accurately by the relation
Yl
^Xj2 = - - 2 (residual error)2.
w — r
Hence if for 2 (residual error)2 we write
Applying these results to Example I on p. 377,
W = 574
n = 7
then ^
* - '6745^ = -2475
MATHEMATICAL PROBABILITY 383
/ 2n^rez /2X574 c
also h = v 7 v == v =i '3°
v (n— i)Xn 6
*. e., h has a very high value ; and this would be expected, for the
"fineness " of reading, as judged by the results, is not at all good
(one error being as much as 2 in 12).
Example 2. — In a chain survey four measurements of a base line
gave 867-35, 867-51, 867-28 and 867-62 links respectively. Find the
best length and the probable error in this length.
The best result is the A.M. of these, i. e.,
867-35+867-5I+867-28+867-62
4
— 867-44 links
and whilst this is the best result it contains a probable error.
Probable error in A.M.
= ^=-6745\/^Yy
= -6745
/(— Q9)2+(-Q7)2+(—
4X3
- -6745 V ^
= -0517
i. e., the base line measurement (867-44) is subject to an error of
•0517 link, and as this result could not be bettered it would be
unnecessary to repeat this portion of the survey.
The probable error in any one observation would be
'='6745 \-- = -io3,
so that there is a decided gain in accuracy obtained by increasing the
number of observations. (Cf. "repetition," when working with the
theodolite.)
It is of interest to find h for this example.
2X-07I _
-J = '2176
and as this is a small quantity we are confirmed in our conclusion that
the observations were well made.
Example 3. — The mean values of the three angles of a spherical
triangle were calculated from the actual observations to be 75° 40' 21 -6",
39° 1 i' 47-3", and 65° 7' 56-2"; and these values were subject to
probable errors 2-9", 3-6*, and 4-3* respectively. From a knowledge
MATHEMATICS FOR ENGINEERS
of the area of the triangle, the spherical excess of the triangle was
found to be 3-3*. Make the necessary adjustments to the angles to
satisfy this condition.
The actual spherical excess
= (75040'2i-6"+390ii/47-3"+65°7/56-2")-i8o0
There is thus (5-1 — 3-3) to be divided among the angles, according to
the respective weights ; and these weights are in the proportion
or -119 -077 -054,
the sum of the weights being -250.
•IIO *O77
Hence the corrections to be applied are Xi-8, — — xi-8. and
•250 -250
— ^ xi '8 to the respective angles; all these corrections being sub-
•250
tracted, since the observed angles give a spherical excess greater than
should actually be the case.
These corrections are -857, -555, and -389.
Hence the true angles are (75° 40' 21-6*— -86*), (39° ii'47'3*— -56")
and (65° / 56-2*- -39*),
or 75° 40' 20-74", 39° II/46'74'/ and 65° 7' 55-81*.
Example 4. — Measurements of an angle in a traverse survey were
made by two different observers, with the following results : —
Readings by A.
Readings by B.
76° 50' 20*
76° 50' 55"
76° 50' 50*
76° 50' 35*
76° 50' 30"
76° 51' 15"
76° 51' 10*
76° 51' 20*
76° 50' 30*
76° 51' o"
76° 51' o"
76° 50' 45*
76° 50' 40*
76° 50' 25*
76° 50' 30*
76° 50' 40"
Compare the two results from the point of accuracy, and find the
most probable value of the angle.
We must first find the arithmetic mean of each set of observations,
and then, by subtracting this from each reading, we determine the
residual errors.
The A.M. of set A = 76° 50' 4 1-25*
and A.M. of set B = 76° 50' 5 1 -88*.
MATHEMATICAL PROBABILITY
385
Since the differences are of seconds only, we need not concern
ourselves for the present with the degrees and minutes ; and thus the
table of residual errors and their squares becomes
Residual Error.
(Residual Error)2.
Residual Error.
(Residual Error)*.
— 21-25 451.4
+ 3-12
9-7
+ 8-75
76-6
— 16-88
284-9
— 11-25
126-6
+ 23-I2
534-6
+28-75
826-8
+ 28-12
790-7
-11-25
.126-6
+ 8-12
66-0
+ 18-75
35 1'6
- 6-88
47-3
- 1-25
1-6
-26-88
722-4
— 11-25
126-6
-H-88
141-2
sum o
2087-8
o
2596-8
In case A f
/
2087-8
»» — '"745 <v
8x7
In case B I
»» °745 V
8x7 "
weight of observations by A (4-594)
2 1-244
(4-119)'
Thus A's readings can be relied on before those of B ; the former
being roughly ij times as good as the latter.
The most probable value of the angle, taking into account the two
sets of readings, will be obtained by the calculation of the " weighted
mean," i. e., the mean of the two arithmetic means already found,
determined with due regard to the respective weights to be given to
A's readings and B's readings.
Dealing only with the seconds, the most probable value
(41-25 x i -244) + (5 1 -88 x i)
1 + 1-244
51-31+51-88 _ 103-19
2-244 2-244
— = — - — - = 46 seconds.
Hence the most probable value of the angle = 76° 50' 46*
Exercises 26. — On the Calculations of Errors of Measurements.
1. One surveying party measured a certain base line as 6 chains
42-7 links, 6 chains 53-5 links, 6 chains 46-4 links, and 6 chains
41-9 links ; and a second party measured the same line as 6 chains 38-4
links, 6 chains 39-7 links, 6 chains 46-9 links, and 6 chains 43 links.
State which of the two parties is the more dependable, and find the
most probable length of the line.
C C
386 MATHEMATICS FOR ENGINEERS
i -*_?
2. Plot the probability curve y = /- e AZ the value of h being
n v TT
•1414, and find the probability that an error lies within the limits
— -6 and + -6.
3. The following are the values of the determination of the azimuth
of Allen from Sears, Texas, the results of a U.S. Coast and Geodetic
Survey; the values of the seconds only being stated after the first
reading: 98° 6'4i-5*, 42-8, 43-4, 43-1, 39-7, 42-7, 41-6, 43-3, 40-0, 45-0,
43-3 and 40-7. Find the A.M., the probable error of a single obser-
vation and the probable error of the mean.
4. Find the weighted mean of the following observations : 95-8,
96-9, 97-2, 95-4, 95'7, 97gii 96-5, 96-7 and 97 ; the probable errors in
the measurements being -2, -4, -i, -9, -7, 1-2, -8, -3 and 1-5 respectively.
ANSWERS TO EXERCISES
Exercises 1
3*7 AC*
3. E = constant x -~ 4. V == RC + L ^
at at
5. —11-03 cms. per sec. : 1-07 seel;, from start 6. -336 ton
11. 5.65 12. Middle of May : middle of October
15. Loading is -2 ton per foot run 16. -966 : — =^— = cos 6
uv
17. -42 ton per foot 21. 6-3
Exercises 2
1.
4#3
2. -128 3. —2* 4. 27*" 5. — g
18-75
•0086 i'ii
-pl.23
#v
1 1
•982
19 -fi?*8'3 13 /<car2 /i/i •Si'1-8 4-
X2'3
X*
14.
0
°* /pi -6 347
17.
•JIV3'Si -
, -84 . 1-29 •52'5 12-48 ._ .-
20.
•073
21. i - a - §&P~* 22. w(^ + Xjr~j-x
23.
•289*
w "* d / I
24. — : (Arv + nyz — xl) 25. « = - \/ i + -;
V/ 2 w
26.
-2(/> +
/*"f \
9) 27.7-85 28. w(-j--x\ 29.9-6
30.
-7333
Exercises 3
1. Sub-normal — 466 ; sub-tangent = i 2. 25-7
3. y = -0256^ ; (# is distance from centre) : -64
+ 142-5 5. M-= — ( -- xy. S = - -: L, — o
6. M = -(--
2\4
7. M = W (l-x) : S = - W : L = o
8. Sub-tangent = — — : sub-normal =
9. 3 10. -~ (wl -wx- aP)
387
388 MATHEMATICS FOR ENGINEERS
Exercises 4
1. -5*~5* 2. 6-I5**'1* 3. -^=| 4. 1-423 (4-15)*
&
5. 4-33 (8-72)2x 6. ge9* — 35e~7x 7. 5-44*1'718 8. 9-7 (2)*
9. io-25<r25*- 10.
11. i2-6e*'2* 12. - 13. ^— 14.
x 5* — 4
4ac^ _!.&,-. 2*
1 O. — r A u. ioe
i^
X1 4
17. I 4 --- ? - or*2L-VT- [Use the rule log AB = logA+logB]
* 3*— 47 *(3*— 47)
18. — -> -- \ -- - -- \ -- ^— 19. 303jc + 8 sinh zx — '^~
5* + 4^3* -2 7-4^ *
20. - X - 1-057 (1-8)* 21. o 22. 1-052 23.
24.^43 25>.2T 26.^,- 27.^7-
t -L* i *ow — 7
28. ^ sinh- :^ cosh y- 29. frE 30. & 31.
44?? ^fc
32. o 33. o 34. - + C - -,
-
Exercises 5
1. — 5'3 cos (4 — 5-3*) 2. —16-3 sin 5-1* 3. -48 sec2 (3* -f 9)
5-05sin(-05 — -117*)
4. -QI4COS (-425^— 1-25) 5. — 4ocosec2# 6. „, — :—
cos2 (-05 — -117*)
7. gbc sin (rf — gx) 8. — 20 sin 5^ — 14 cos (zx — 5)
9. 4-40038-8^+ -S cos 1-6^ {Usetherule : 2 sinAcosB = sin(
-(- sin (A — B)} 10. — 6-74 sin 6-2X — 3-04 sin 2-8
11. 4'52 V ~^ sin (Px — 1X + 2C) + ., sin (px
\O 'U "U ™T" U
12. 5 sin 2x. {Use the rule: cos 2A = i — 2 sin2 A} 13. —-195 sin 6x
14. o 15. S-I6A"72 -5-2 -0273 cos (4-31 — -195^) + 24-93
3# — 4'1
16. -1056 cos -015^ — -0529 sin (6-1 — -23*) + 7-4 sec2 (4* — -07)
17. Velocity = 37-7 sin 31-4* — 56-56 cos 31-4* :
acceleration =1184 cos 31-4/4- J777 sin 3I-4'
18. Acceleration —— -02895 : S.H.M. 19. —1162
20. Sine curve (i.e., second derived curve).
f B/
21. o I Treat as a constant the portion
22. — — 1500^ cos pt + $oop cos $pt + 42^ sin pt — 8^p sin
ANSWERS TO EXERCISES 389
Exercises 6
1. 2 cos 2.x . <?8in2* 2. - 3» — 2 sin 2* 4. 24*2. cos x3
v
5. 3-14 (iox + 7) sec2 (5#2 + 7* — 2) 6. 3 log a . cos 3* . a"in3*
3 ~T" i~j % ~~~~ Q# 5
sec20 ._ sin<9 (dy dy dd\
10. COSCCA- 11. 12. - —. { — = 3^ - X 3
secz cp cos a i + cosy la^r aw a^J
13. \ ,, ,j. — slope of curve = — j^— X -=- X — ^ — =- ,
IrflogV rfA dV rflogV)
15. i -08 ft. per min. : -377 ft. per min. 18. -033215.: •01020'
19. 7 cosec 7* + 45*2 20.=^ 21. 56° 19' 22. 53° 7' 24.
Exercises 7
1. x (2 sin 3* -f $x cos 3*) 2. 2^2'4 (i + 3-4 log
3.
( 2-07 5 sin (3-1- 2-07*) \
Icos -1 — 2-0^ x J
4
Ar5 cos (3-1 — 2-07.*) cos (3-1 — 2-07^)
5. - {2-575 sin (5-15* + 4) + -625 sin (1-25* - 4)} or
- {3-2 sin 3-2* cos (i-95* + 4) + i'95 cos 3-2^ sin (1-95* + 4)}
6. sec2 2x {2 cos (5 — 3#) + i -5 sin (5 — 3*) sin 4*} or
3 tan 2.x sin (5 — 3*) + 2 sec2 2# cos (5 — 3*)
7. I2-8*'6 (cos (3 + 8x) + 2 — 5* sin (3 + 8*)}
8. 27 (5)3z J4-83 log x + - 9. (i + log AT) exlo«x 10.
11. 3oe— (5* + 4)(5* + 2) 12. /tan -125^ log^ \
3 ^\ x 8cos2-i25Ar/
13. o 14. — 5e-lot 15. o 16. 12600 sin (14*— -116)
17. 6t {5 sin (4 — -8t) — 2t cos (4 — -8*)) 18. 4*2'7 (3-7 cos 3^ — 3* sin 3*)
Exercises 8
1. 5*2(73~7;y) 2. - -. (-^— - log (2 - 7*) tan (2 - 7*)}
e7*-5 cos (2— 7#) \7# — 2
3 20 4 & 5 5-46 (5)2^
'"
Vd* —
6. TCsin^n-^-^S^coshi-S^ or
— 21 i —
_
*2 + 6^ + 15 (i -
Q wb (ab — -zbx + x2 cot B)
~2~(&-*cotB)2
39o MATHEMATICS FOR ENGINEERS
es\n(i-2x + l-7)
12< {loi^^T^Ts)? x
j I -2 COS (l-2X -{- I -7) log (8xz — JX + 3) + -~-
. 6-55 (sin 0 — 0 cos 0)
13. sech2* 14. J3.v
05 (0 — sin 0)^
. 66-2 (0 — sin 0)* (20 — 30 cos 0 + sin 0) 4 „ yz
15. lo.
17. rw2 cos 0 -1 —
L (m2 - sin2 0)*
d(f> a> cos 0 . dz(f> _ a>z sin 0 (i — m2)
'' ~dt ~ \/m2 - sin2 0 ' ^2 (ma-sina0)^
19. ^^a: ± J* 20.
(# — ?)sm 2^ > w
Exercises 9
2.
*pt + 4e« . log (5p - 3)
4. io(4-w)(9-4w)(3 + 8w)2 6. 8(1-7
?_(*__
' CrVdt r'dt
7 _ 31 o ?_(*P__t dr\
°' '
Exercises 10
1. 750 2. 17-1 3. ^ : ^- 4. -577/ : -I28W/
5. -5 6. — 2-25 : minimum 7. -278 : maximum
8. maximum at x = —5 : point of inflexion at x = —2
minimum at x = i 9. 2 rows of 8
10. base = 3-652 ft. : height = 1-826 ft. 11. 2-1 : minimum
12. - 13. -496: £631 14. width = height = 8-4'
15. 15-2 knots: £956, £948, £957 16. x = -289;
17. h = 6-34 ft. : d = 12-68 ft. 18. base = 4* : height = 5*
19. depth = 3 x breadth. 20. ^ 21. 6 22. -866 r
w
23. v : | : — 24. u = -.51; 25. 135° or 315°
ANSWERS TO EXERCISES 391
26. d = l 27. I = 2'o6sr 28. 20-15: —-45 29. o or 20° 56'
30. height = 8-1 ft. : base = 6-72 ft. 31. x = -4*
32. tan- (-*±^?±$\ 38. / _
i — fiz
34. maximum at x = o : points of inflexion at x —• ± —j=
. V2
35. VP 36. /(4/+d 37.
K
i + K
Let r = ^ and find ^] 39. 3 units 40. * = ^^
P! dr J 2
41. T/ = ^Tm 42. maximum at * = — 2
minimum at * = 4 and at * = —2-5
points of inflexion when * = —2-26 and 1-92
/
38. -58 (
\
43. x = VRjRg : /»« (maximum) = - fam + i) (Rx — R2)2
44. d = \/^j 46. 83° i' or 276° 59'
' ojt
Exercises 11
1. -006 2. 2«5%toolow 3. -0264 4. —2-45
5. 2-66 6. decrease of -00135 7. -03 link; -237 link
, (#loga)2 , (#loga)3 ,
8. i + x log a + * - ! :— i- + . . .
1.2 1.2.3
9. 2214-2525 10. -536
Exercises 12
2. 152 5. 240 ; 205 ; 64
6. 621,000 ft. Ibs. : potential energy = 240,000 :
kinetic energy = 381,000 : 987,000 ft. Ibs.
7. 480 9. 238,000 10. 3006 12. i-526#2-68+C
13. 70-15* + C 14. - +C 15.
17. -i*10 — 10 log x + 14* + C 18. 3-32*1'04 — 2-5*2 + C
19. i-o74*3'718 + ie* + C 20. i-33«9*-f + C
•
21. 6-54««-*'-1-» + C 22. -689* + C 23. -^ - + C
C
24. 3-025*'84 — 8-2 log x — 2-7ie-2>«* + 1-13* + C
25. — -0234*-10'2* + C 26. i-g6e'61x — 1-297*-" + -674*8'04 + C
27. -797 cos 0*1-" — 2-2g-8'-i*+C 28. Ju6 + C 29. -- ^§+C
30. 35< + C 31. ie^-«+C 32. - 5-88^ + C
33. 20-2 (2)» + C 34. - * + 2-5^2 + 4-25** - 8x + C
^T
35. -885(3-i)t + C 36. _ + C
392 MATHEMATICS FOR ENGINEERS
37. _ _ z-jiSt + C 38. i6-i*2 + C 39. - + C
45 x
40. Write the equation in the form -f — — w — and then integrate :
p v
pvn — C
Exercises 13
1. -Jcos4* + C 2. i -73 sin (3 - 3*) + C
3. — 49 tan (3 — }x) + C 4. i-oix'9** + 1-195 sin (-05 — -117^) + C
5. -I85405'4*— -cos (b + ax) + C 6. 9-45* . sin 8t + C
&
7. -713 cos 2(2-i6x — 4-5) + C
8. i2-85e'7* — ^— r + i -83 log cos # + C
A# — 2-8\
9. —9-95 cos (— ) + '022 sin 9^r — i'46^2'74 + -455(3)2x + 6 + C
\ 7 /
10. 2* — -787 cos ( - — 3-7* ] + 7-55 cot. ^ + C
V4 / 5
(i
11. v = 7 cos (7^ — -26) + C ; s =
49
12. -5^ = 47r2nV( sin ^ + Sin 2 \ + C; * = — 47r2nV( cos ^ + C°S 2 ) + C
«D \ 2m / \ 4m /
13. •3iH58p) - -139 sin (3-7 - 7-2^) + C
14. — 19-5 cos 6t — 4-9 sin 6t + C
Exercises 14.
1. -182 2. -345 3. 1-7 4. -561 5. -0626
2!
6. 1-218 x io7 7. 2-62 8. -1589 9. -
P
10. -6i6Bmai 11. - - |C— (aJ"^ + C^S (fl ~ -f W
2 I a + 6 a — 6
^o -n
2 13- p =
14. » {^4 - 4/3^ + 3,4} 1 -
24EI (
16. ^ 17. .2046/* 18. I^SiU^ • Wk*
3° i + sin $ 2
19. 26-24 {Limits must first be found} ; f 20. 334
21. y = -736*?'« + 5 log # + 3^ — 3-25 22. i -087.
. . ._
26.H = g 27.240 3 ;8.,9I 29.8
30. 49-82 31. ^f 32. -8596CA.W 33.
ANSWERS TO EXERCISES 393
Exercises 15
1. -158 log - 2.
5 (* +1-583) 3V6 V6
3. log (9*2 - 18* + 17)" -f C 4. -1919
5. 605 {Let u — 6 — h] 6. yV-j$ — y2 + -75 sin"1 1 -154^ + C
7. -106 8. $sini2* + 4/ + C 9. — tcot5#+C
.. WR3/i i\ -0683 WR3
10. -==- --- ) or -- £- —
El \ir 47 El
11. — -8 log cos 5* + C {Let u = cos 5*} 12. # sin"1 # + Vi — ^ • + C
_ -g&Z
13. — £V (a2 — AT2)3 + C 14. 3 — {3 sin 2* — 2 cos -zx]
15. tan-1 * H -- ~ + C (Let u - tan-1 AT) 16. -081
i + xz
17 tsin»<9 18
18> k
19. 183 sees. : {Rationalise denominator of right hand side by
multiplying top and bottom by Vh + 12 — Vh ; then integrate, making
the substitution u = A + 12.}
21.
10 42
22. ^ + t - 1 cosh-1 1 + C 23. 355 24. i -749
12 8 2 2 256
25. — 26. -oiR3
315
27. (-5* + 1-25) 2T^5* -x* + 13-63 sin-1 + C
28. F=
V a2
Exercises 16
1.. 41-59 2. 1-718 3. 10-85 4- <688 5- 4'5
6. 23-05 7. 1-348 ft. candles. 8. o 9. -1294
10. 273 11. 205 Ibs.
Exercises 17
1. 14-14 2. -6215 3. 2-4 4. -1165 5. -833 6. 1-194
394 MATHEMATICS FOR ENGINEERS
Exercises 18
1. 1450 cu. yds. 2. 5977 Ibs. 3. 1070 4. 4nr2
c KTrD2/2 x_ . Kl
5. a — 3-036, b = -1423 ; 1617 6. — ^ — or Vol. x
7. 524 (limits are 5 and 10} 8. 271-6 9. 1-2 Ibs.
10. true == 76-62 : (a) 75-41 : 1-58 % low (b) 77-73 : 1-45 % high
(c) 76-60 : correct. 11. 60-9 ft.
Exercises 19
1. | height from vertex
2. area = 7200 sq. ft. : centroid is 158' from forward end
3. 3-84 Ibs. ; 4" 4. 10 ft. from top
5. x = y = i -7* (taking the centre as origin) 6. 771 ; 2-25
7. (o, -95) 8. 2-35* from AB 9. 1-055* 10. -935"
11. -877* 12. 1-02* 13. 2-68* i. e., 1-34*
14. 5-12 ft. from O 15. 30880 Ibs. 16. 18432 Ibs. : 3-534 ft.
17. 9 units. 18. 5-1 ins. 19. 2-35 ins.
20. (a) 17 Ibs. (b) 1 1* 21. 1-48 ins. * 22. \h 23. |A
Exercises 20
1. -655/ 2. 17-11 ins.: 785 Ibs. ins.2
3. C. of G. is -1125* distant from centre of large circle: 2-68"
4. 16-6 ins.4 5. (a) -408^ ; (6) -707^
6. IAB = 80-7 ins.4 : &AB = 3'O4* 7. 29-1* 8. 377
9. 9-86 10. 7-35
11. -2887^. (Divide into strips by planes perpendicular to the axis
and sum the polar moments of these)
. IE of circular 3
12. T_ , - = - = -956 13. 33.3 inch units
IE of square TT
14. 681-6: 17-1
15. INN = 169-4: £NN = 2-44: IAB = 570 : k^ = 4-47
16. NN is 3-99* from bottom of lower flange : INN = 461 : £NN = 3'77
17. 5-04* 18. 2-023': -444 19. 2-74*
20. (a) 13-9 (6) 31-1 (c) 1-48" (d) 1-82* 21. -28
22. massR* + 23. -56
Exercises 21
3. 5-23 4. £ = 2rsin0: the sine curve 6. 892 7. 5-01
Exercises 22
1. y = i-6jx* — 2-4* — 12-82 2. s = 8-05^ — 23-1* + 14-1
3. y = Ae** — £ 4. y = 8-35 — -1490 -i-o*«*
W fdv I
5. y = - 2 — 3 — 3 = = = = -
ANSWERS TO EXERCISES 395
wlz w
7. log = - 8. v = (5« -
9- l°g = 10.
ws
11. H =
,— 7-2) . / ftnh
J — 12. 0 = Asm(/ -
,0 , B
13. ^ = A+-2
17. y = A^10-6 + Ajg7' : y = A^10' + A^7* + i
18. 5 = A^9'33' + A^-9'33' 19. 5 = A sin (9-33^ + B)
_
20. KRlog-1 21. C = C0Ae 22. C=
23. ^ = alo or a=
(Separate the variables and use the / -5 - ^ form J
Ysin /y/ ==,
25. =- -
26. Mf = log^+ — -):-i945 27. AT=
28. x = Ae~3t sin (6«32/ + c) + -026 a sin (5^ — tan-1 1-25)
J\ - /5.
29. V - Axe V r* + A2e V r* 30. /w» - C
31. 6 = 6a + e-kt(60-6a} 32. 5= - 53-636-^^+ 5-83e
where
=
•V
V
2EI
e
2EI
or y = B^* + B2«-« -f B3 sin <f>x + B4 cos (f>x — -^
34. « = '
396 MATHEMATICS FOR ENGINEERS
Exercises 23
1. 103 sees, if coefficient of discharge is taken as -62 2. 7-37
3. 2-83 cu. ft. 4. 69-5 : 95-5 (Draw in the " simply supported "
bending moment diagram and work on the Goodman plan, see page 313)
5. Find the time to lower to level of upper orifice (183 sees.) with
both orifices open; then the time for the further lowering of 5 ft.,
through the one orifice (180 sees.). Total time = 363 sees. Note that
*
6. 57-7 sees. 7. 1-4
12
8. 5500 Ibs. : 4-71 ft. below S.W.S.L. 9. 14100 Ibs. : 6-65'
below S.W.S.L. 10.761 11. '^f® G;.- ,^),) (Hint.
let u = de+Kx) 12.1-23 13. Vertical depth = 8-07 ft.
Exercises 24
1. x = 2-31 — 1-231 cos 5— 1-55 sin 6— -16 cos 20
— -022 sin 20— -004 cos 3#— -04 sin 3$
2. A— 1-29, oj = O, B = -I4, o2 = rr
3. ^ = 16-97+6-49 cos #+ -002 sin x— 12-66 cos 2*
— i -46 sin 2.x— 1-75 cos 3#— -7 sin $x
4. E = 1500 sin 0+too sin 30—42 cos #+28 cos 3$
Exercises 25
1. B = 39°3i/; A = 65°5i'; c = 57° 5'
2. c - 54° 44i' ; * = 34° 14' •' B = 43° 32^'
3. A= 161° 8'; B= 13° 35'; C = 9° 38'
4. A = 76° 36'; B = 64°8'; 0 = 52°
5. B = 35° 43' 40"; A = 6i° 21' 20"; a = 43° 25' 23"
6. C= 33° 29' or 146° 3 1'
a = 103° 28' or 55° 28'
A= 146° 58' or 27° 30' 7. 55° 10' 8. 342,200
9. Latitude = 43° 54' : hour angle = 4° 31-3'
Exercises 26
1. 2nd set better than ist set in the proportion 1-943 to i : 6 elms.
43-4 links. 2. Just under i. 3. 98° 6' 42-26" : ym — -307", v — 1-062"
4. 96-93
MATHEMATICAL TABLES
TABLE I. — TRIGONOMETRICAL RATIOS
Angle.
Chord.
Sine.
Tangent.
Co- tangent.
Cosine.
De-
grees.
Radians.
0°
o
o
0
o
GO
i
1-414
1-5708
90°
i
2
3
4
•0175
•0349
•0524
•0698
•017
•035
•052
•070
•0175
•0349
•0523
•0698
•0175
•0349
•0524
•0699
' 7'290O
28-6363
I9-08II
I4-3007
•9998
•9994
•9986
•9976
1-402
1-389
1-377
1-364
1-5533
1-5359
1-5184
1-5010
89
88
87
86
5
•0873
•087
•0872
•0875
II-43OI
•9962
l'35l
I-4835
85
6
8
9
•1047
•1222
•1396
•1571
•105
•122
•I40
•157
•1045
•1219
•1392
•1564
•1051
•1228
•1405
•1584
9-5I44
8-H43
7-H54
6-3I38
•9945
•9925
•9903
•9877
1-338
1-325
1-312
1-299
1-4661
1-4486
1-4312
1-4137
84
83
82
81
10
•1745
•174
•1736
•1763
5-67I3
•9848
1-286
1-3963
80
ii
12
13
14
15
•1920
•2094
•2269
•2443
•192
•209
•226
•244
•1908
•2079
•2250
•2419
•1944
•2126
•2309
•2493
5-I446
4-7046
4-33I5
4-OIO8
•9816
•9781
•9744
•9703
1-272
1-259
1-245
1-231
1-3788
1-3614
1-3439
1-3265
79
78
77
76
•2618
•261
•2588
•2679
3-7321
•9659
1-218
1-3090
75
16
17
18
19
•2793
•2967
•3*42
•33i6
•278
•296
•313
•330
•2756
•2924
•3090
•3256
•2867
•3°57
•3249
•3443
3-4874
3-2709
3-0777
2-9042
•9613
•9563
•95"
•9455
1-204
1-190
1-176
1-161
1-2915
1-2741
1-2566
1-2392
74
73
72
7i
20
•3491
•347
•3420
•3640
2-7475
•9397
1-147
1-2217
70
21
22
23
24
•3665
•3840
•4014
•4189
•364
•382
•399
•416
•3584
•3746
•3907
•4067
•3839
•4040
•4245
•4452
2-6051
2-4751
2-3559
2-2460
•9336
•9272
•9205
•9135
1-133
1-118
1-104
1-089
1-2043
1-1868
1-1694
1-1519
69
68
67
66
25
•4363
•433
•4226
•4663
2-1445
•9063
1-075
I-I345
65
26
27
28
29
•4538
•4712
•4887
•5061
•45°
•467
•484
•501
•4384
•454°
•4695
•4848
•4877
•5095
•5317
•5543
2-0503
1-9626
1-8807
1-8040
•8988
•8910
•8829
•8746
i -060
1-045
1-030
1-015
1-1170
1-0996
1-0821
1-0647
64
63
62
61
30
•5236
•518
•5000
•5774*
I-732I
•8660
I-OOO
1-0472
60
31
32
33
34
•54ii
•5585
'•5760
•5934
•534
•551
•568
•585
•5150
•5299
•5446
•5592
•6009
•6249
•6494
•6745
1-6643
1-6003
1-5399
1-4826
•8572
•8480
•8387
•8290
•985
•970
•954
•939
1-0297
1-0123
•9948
•9774
59
58
57
56
35
•6109
•601
•5736
•7002
1-4281
•8192
•923
•9599
55
36
37
38
39
•6283
•6458
•6632
•6807
•618
•635
•651
•668
•5878
•6018
•6i57
•6293
•7265
•7536
•7813
•8098
1-3764
1-3270
1-2799
1-2349
•8090
•7986
•7880
•7771
•908
•892
•877
•861
•9425
•9250
•9076
•8901
54
53
52
51
40
•6981
•684
•6428
•8391
1-1918
•7660
•845
•8727
5°
41
42
43
44
•7156
•7330
•7505
•7679
•700
•717
•733
•749
•6561
•6691
•6820
•6947
•8693
•9004
•9325
•9657
1-1504
1-1106
1-0724
1-0355
•7547
•7431
•7314
•7193
•829
•813
•797
•781
•8552
•8378
•8203
•8029
4i
48
47
46
45°
•7854
•765
•7071
I-OOOO
I-OOOO
•7071
•765
•7854
45°
Cosine
Co-tangent
Tangent
Sine
Chord
Radians
Degrees
Angle
397
398
MATHEMATICAL TABLES
TABLE II.— LOGARITHMS
J
0
1
2
3
4
5
6
7
8
9
123
456
789
10
0000
0043
0086
0128
0170
0212
0253
0294
0334
0374
4 9 13
4 8 12
17 21 26
16 20 24
30 34 38
28 32 37
11
0414
0453
0492
0531
0569
0607
0645
0682
0719
0755
4 8 12
4 7 11
15 19 23
15 19 22
27 31 35
26 30 33
12
0792
0828
0864
0899
0934
0969
1004
1038
1072
1106
3 7 11
3 7 10
14 18 21
14 17 20
25 28 32
24 27 31
13
1139
1173
1206
1239
1271
1303
1335
1367
1399
1430
3 7 10
3 7 10
13 16 20
12 16 19
23 26 30
22 25 29
14
1461
1492
1523
1553
1584
1614
1644
1673
1703
1732
369
369
12 15 18
12 15 17
21 24 28
20 23 26
15
1761
1790
1818
1847
1875
1903
1931
1959
1987
2014
369
368
11 14 17
11 14 16
20 23 26
19 22 25
16
2041
2068
2095
2122
2148
2175
2201
2227
2253
2279
358
358
11 14 16
10 13 15
19 22 24
18 21 23
17
2304
2330
2355
2380
2406
2430
2465
2480
2504
2529
368
267
10 13 15
10 12 15
18 20 23
17 19 22
18
2553
2677
2601
2625
2648
2672
2695
2718
2742
2766
267
257
9 12 14
9 11 14
16 19 21
16 18 21
19
2788
2810
2833
2856
2878
2900
2923
2945
2967
2989
247
246
9 11 13
8 11 13
16 18 20
15 17 19
20
3010
3032
3054
3075
3096
3118
3139
3160
3181
3201
246
8 11 13
15 17 19
21
22
23
21
3222
3424
3617
3802
3243
3444
3636
3820
3263
3464
3655
3838
3284
3483
3674
3856
3304
3502
3692
3874
3324
3522
3711
3892
S345
3541
3729
3909
3365
3560
3747
8927
3385
3579
3766
3945
3404
3598
3784
3962
246
246
246
245
8 10 12
8 10 12
7 9 11
7 9 11
14 16 18
14 15 17
13 15 17
12 14 16
25
£979
3997
4014
4031
4048
4065
4083
4099
4116
4133
235
7 9 10
12 14 16
26
27
28
29
4150
4314
4472
4624
4166
4330
4487
4639
4183
4346
4502
4654
4200
4362
4518
4669
4216
4378
4533
4683
4232
4393
4548
4698
4249
4409
4664
4713
4265
4425
4579
4728
4281
4440
4594
4742
4298
4456
4609
4757
235
235
235
134
7 8 10
689
689
679
11 13 15
11 13 14
11 12 14
10 12 13
30
4771
4786
4800
4814
4829
4843
4867
4871
4886
4900
134
679
10 11 13
31
32
S3
34
4914
5051
5185
5315
4928
5065
6198
5328
4942
5079
6211
5340
4955
5092
5224
5353
4969
5105
6237
5366
4983
5119
5250
5378
4997
6132
5263
6391
6011
5145
6276
6403
6024
5159
5289
6416
6038
5172
6302
6428
134
134
134
134
678
678
568
568
10 11 12
y 11 12
9 10 12
9 10 11
36
5441
5453
6465
5478
5490
6502
6514
6527
5639
6551
124
567
9 10 11
86
37
38
39
5563
5682
5798
6911
5576
5694
5809
5922
5587
5705
5821
5933
5599
6717
5832
5944
6611
6729
6843
5955
6623
6740
5855
5966
5635
6752
5866
6977
6647
6763
6877
6988
5658
6775
5888
5999
5670
5786
5899
6010
124
123
123
123
667
667
667
457
8 10 11
8 9 10
8 9 10
8 9 10
40
6021
6031
6042
6053
6064
6076
6086
6096
6107
6117
123
466
8 9 10
41
42
43
44
6128
6232
6335
6435
6138
6243
6345
6444
6149
6253
6355
6454
6160
6263
6365
6464
6170
6274
6375
6474
6180
6284
6385
6484
6191
6294
6395
6493
6201
6304
6405
6603
6212
6314
6416
6513
6222
6325
6425
6522
123
123
123
123
456
456
456
456
789
789
789
789
45
6532
6542
6551
6561
6671
6580
6590
6599
6609
6618
123
456
789
46
47
48
49
6628
6721
6812
6902
6637
6730
6821
6911
6646
6739
6830
6920
6656
6749
6839
6928
6665
6758
6848
6937
6675
6767
6857
6946
6684
6776
6866
6955
6693
6785
6875
6964
6702
6794
6884
6972
6712
6803
6893
6981
123
123
123
123
4 5 <
465
445
445
778
678
678
678
50
6990
6898
7007
7016
7024
7033
7042
7060
7059
7067
1 2 S
345
678
MATHEMATICAL TABLES
TABLE II. (contd.)
399
0
1
2
3
4
5
6
7
8
9
123
456
789
51
52
53
54
7076
7160
7243
7324
7084
7168
7251
7332
7093
7177
7259
7340
7101
7185
7267
7348
7110
7193
7275
7356
7118
7202
7284
7364
7126
7210
7292
7372
7135
7218
7300
7380
7143
7226
7308
7388
7152
7235
7316
7396
123
122
122
122
345
345
345
345
678
677
667
667
55
7404
7412
7419
7427
7435
7443
7451
7459
7466
7474
122
345
567
58
67
58
59
7482
7559
7634
7709
7490
7566
7642
7716
7497
7574
7649
7723
7505
7582
7657
7731
7513
7589
7664
7738
7520
7597
7672
7745
7528
7604
7679
7752
7536
7612
7686
7760
7543
7619
7694
7767
7551
7627
7701
7774
122
122
112
112
345
345
344
344
567
567
567
567
60
7782
7789
7796
7803
7810
7818
7825
7832
7839
7846
112
344
566
61
62
63
64
7853
7924
7993
8062
7860
7931
8000
8069
7868
7938
8007
8075
7875
7945
8014
8082
7882
7952
8021
8089
7889
7959
8028
8096
7896
7966
8035
8102
7903
7973
8041
8109
7910
7980
8048
8116
7917
7987
8055
8122
112
112
112
112
344
334
334
334
566
566
& 5 6
556
65
8129
8136
8142
8149
8156
8162
8169
8176
8182
8189
112
334
556
66
67
68
69
8195
8261
8325
8383
8202
8267
8331
8395
8209
8274
8338
8401
8215
8280
8344
8407
8222
8287
8351
8414
8228
8293
8357
8420
8235
8299
8363
8426
8241
8306
8370
8432
8248
8312
8376
8439
8254
8319
8382
8445
112
112
112
112
$ 3 4
334
334
234
556
556
456
456
70
8451
8457
8463
8470
8476
8482
8488
8494
8500
8506
1 1 2
234
456
71
72
73
74
8513
8573
8633
8692
8519
8579
8639
8698
8525
8585
S645
8704
8531
8591
8651
8710
8537
8597
8657
8716
8543
8603
8663
8722
8549
8609
8669
8727
8555
8615
8675
8733
8561
8621
8681
8739
8567
8627
868G
8745
112
112
112
112
234
234
234
234
455
455
455
455
75
8751
8756
8762
8768
8774
8779
8785
8791
8797
8802
112
2 3 3
455
76
77
78
79
880S
8865
8921
8976
8814
8871
8927
8982
8820
8876
8932
8987
8825
8882
8938
8993
8831
8887
8943
899S
8837
8893
8949
9004
8842
8899
8954
9009
8848
8904
8960
9015
8854
8910
8965
9020
8859
8915
8971
9025
112
112
112
112
2 S 3
233
233
233
455
445
445
445
80
9031
9036
9042
9047
9053
9058
9063
9069
9074
9079
112
233
446
81
82
83
84
9085
9138
9191
9243
9090
9143
9196
9248
9096
9149
9201
9253
9101
9154
9206
9258
9106
9159
9212
9263
9113
9165
9217
9269
9117
9170
9222
9274
9122
9175
9227
9279
9128
9180
9232
9284
9133
9186
9238
9289
112
112
112
112
233
233
233
233
446
446
446
445
65
9294
9299
9304
9309
9315
9320
9325
9330
9335
9340
112
233
445
86
87
88
89
9345
9395
9445
9494
9350
9400
9450
9499
9355
9405
9455
9504
9360
9410
9460
9509
9365
9415
9465
9513
9370
9420
9469
9518
9375
9425
9474
9523
9330
9430
9479
9528
9385
9435
9484
9533
9390
9440
9489
9538
112
Oil
Oil
Oil
233
223
223
223
445
344
344
344
90
9542
9547
9552
9557
9562
9566
9671
9576
9581
9586
Oil
223
344
91
92
93
94
9590
9638
9585
9731
9595
9643
9689
9736
9600
9647
9694
9741
9605
9(352
9699
9745
9609
9657
9703
9750
9614
9661
9708
9754
9619
9666
9713
9759
9624
9671
9717
9763
9628
9675
9722
9768
9633
9680
9727
9773
Oil
Oil
Oil
Oil
223
223
223
223
344
344
344
344
95
9777
9782
9786
9791
9795
9800
9805
9809
9814
9818
Oil
223
344
96
97
95
99
9823
9868
9912
995
9827
987-;
9917
9961
9832
9877
9921
99C5
9836
9881
9926
9969
9841
9886
9930
9'J74
9845
9890
9934
9978
9850
9894
9939
9983
9854
9899
9943
9987
9859
9903
9948
9991
9863
9908
9952
9996
Oil
Oil
Oil
Oil
223
223
223
223
344
344
344
334
400
MATHEMATICAL TABLES
TABLE III. — ANTILOGARITHMS
0
1
2
3
4
5
6
7
8
9
123
456
789
•00
1000
1002
1005
1007
1009
1012
1014
1016
1019
1021
001
111
212
•01
•02
•03
•04
1023
1047
1072
10-J6
1026
1050
1074
1099
1028
1052
1076
1102
1030
1054
1079
1104
1033
1057
1081
1107
1035
1059
1084
1109
1038
1062
1086
1112
1040
1064
1089
1114
1042
1067
1091
1117
1045
1069
1094
1119
001
001
001
Oil
111
111
ill
112
222
222
222
223
•05
1122
1125
1127
1130
1132
1135
1138
1140
1143
1146
Oil
112
222
•06
•07
•03
•09
1143
1175
1202
1230
1151
1178
1205
1233
1153
1180
1208
1236
1156
1183
1211
1239
1159
1186
1213
1242
1161
1189
1216
1245
1164
1191
1219
1247
1167
1194
1222
1250
1169
1197
1225
1253
1172
1199
1227
1256
Oil
Oil
Oil
Oil
112
112
112
112
222
222
223
223
•10
1259
1262
1265
1268
1271
1274
1276
1279
1282
1285
Oil
112
223
•11
•12
•18
•14
1288
1318
1349
1380
1291
1321
1352
1384
1294
1324
1355
1387
1297
1327
1358
1390
1300
1330
1361
1393
1303
1334
1365
1396
1306
1337
1368
1400
1309
1340
1371
1403
1312
1343
1374
1406
1315
1346
1377
1409
Oil
Oil
Oil
Oil
122
122
122
122
223
223
233
233
•15
1413
1416
1419
1422
1426
1429
1432
1435
1439
1442
Oil
122
233
•16
•17
•18
•19
1445
1479
1514
1549
1449
1483
1517
1552
1452
1486
1521
1556
1455
1489
1524
1560
1459
1493
1528
1563
1462
1496
1531
1567
1466
1500
1535
1670
1469
1503
1638
1674
1472
1607
1542
1578
1476
1510
1545
1581
0 1 1
Oil
Oil
Oil
122
122
122
122
233
233
233
333
•20
1685
1589
1592
1596
1600
1603
1607
1611
1614
1618
0 1 1
1 2 2
333
•21
•22
•23
•24
1623
1660
1698
1738
1626
1663
1702
1742
1629
1667
1706
1746
1633
1671
1710
1750
1637
1675
1714
1754
1641
1679
1718
1758
1644
1683
1722
1762
1648
1687
1726
1766
1652
1690
1730
1770
1656
1694
1734
1774
0 1 1
0 1 1
Oil
0 1 1
222
222
222
222
333
333
334
334
•25
1778
1782
1786
1791
1795
1799
1803
1807
1811
1816
Oil
222
334
•26
•27
•28
•29
1820
1862
1905
1950
1824
1866
1910
1954
1828
1871
1914
1959
1832
1875
1919
1963
1837
1879
1923
1968
1841
1884
1928
1972
1845
1888
1932
1977
1849
1892
1936
1982
1854
1897
1941
1986
1858
1901
1945
1991
Oil
Oil
Oil
Oil
223
223
223
223
334
334
344
3 4 i
•30
1995
2000
2004
2009
2014
2018
2023
9028
2032
2031
Oil
223
344
•31
•82
•33
•34
2042
2089
2138
2188
2046
2094
2143
2193
2051
2099
2148
2198
2058
2104
2163
2203
2061
2109
2158
2208
2065
2113
2163
2213
2070
2118
2168
2218
2076
2123
2173
2223
2080
2128
2178
2228
2084
2133
2183
2234
Oil
Oil
Oil
112
223
223
223
233
344
344
344
446
•35
2239
2244
2249
2254
2259
2265
2270
2275
2280
2286
112
233
446
•36
•37
•33
•39
2291
2344
2399
2455
2296
2360
2404
2460
2301
2355
2410
2466
2307
2360
2415
2472
2312
2366
2421
2477
2317
2371
2427
2483
2323
2377
2432
2489
2328
2382
2438
2495
2333
2388
2443
2500
2339
2393
2449
2506
112
112
112
112
233
233
233
233
445
446
446
455
•40
2512
2518
2523
2529
2535
2541
2547
2553
2559
2564
112
234
455
41
•42
•43
•44
2570
2630
2692
2754
2576
2636
2698
2761
2582
2642
2704
2767
2588
2649
2710
2773
2594
2655
2716
2780
2600
2661
2723
2786
2606
2667
2729
2793
2612
2673
2735
2799
2618
2679
2742
2805
2624
2685
2748
2812
112
112
112
112
234
234
334
334
455
456
456
466
•45
2818
2825
2831
2838
2844
2851
2858
2864
2871
2877
112
334
656
•46
•47
•48
•49
2884
2951
3020
3090
2891
2958
8027
8097
2897
2966
3034
3105
2904
2972
3041
3112
2911
2979
3048
3119
2917
2985
3055
S126
2924
2992
3062
3133
2931
2999
3069
3141
2938
3006
3076
3148
2944
3013
3083
3155
112
112
112
112
334
334
344
344
566
666
666
666
MATHEMATICAL TABLES
TABLE III. (contd).
401
0
1
2
3
4
5
6
7
8
9
123
456
789
•50
3162
3170
3177
318
3192
3199
3206
3214
3221
322
112
344
5 6 t
•5
•5
•53
•54
3236
3311
3388
3467
3243
3319
3396
3475
3251
3327
3404
3483
3258
3334
3412
3491
3266
3342
3420
3499
3273
3350
3428
3508
3281
3357
3436
3516
3289
3365
3443
3524
3296
3373
3451
3532
3304
338
345
3540
122
133
122
122
345
346
345
345
567
667
667
667
•55
3548
3556
3565
35J3
3581
3589
3697
3606
3614
3622
122
346
677
•56
•57
•58
59
3G31
3715
3802
3890
3639
3724
3811
3899
3648
3733
3819
3908
3656
3741
3828
3917
3664
8750
3837
3926
3673
3758
3846
3936
3681
3767
3855
3945
3690
3776
3864
3954
3698
3784
3873
3963
3707
3793
3882
3972
123
123
123
123
346
346
445
455
678
678
678
678
•60
3981
3990
3999
4009
4018
4027
4036
4046
4055
4064
1 2 3
456
*,,
•61
•62
•63
•64
4074
4169
4266
4365
4083
4178
4276
4375
4093
4188
4285
4385
4102
4198
4295
4395
4111
4207
4305
4406
4121
4217
4315
4416
4130
4227
4325
4426
4140
4236
4335
4436
4150
4246
4345
4446
4159
4256
4355
4457
123
123
123
123
466
466
466
456
789
789
789
789
•65
4467
4477
4487
4498
4508
4519
4629
4539
4550
4560
123
456
789
•66
•67
•68
69
4571
4677
4786
4898
4581
4688
4797
4909
4592
4699
4808
4920
4603
4710
4819
4932
4613
4721
4831
4943
4624
4732
4842
4955
4634
4742
4853
4966
4645
4753
4864
4977
4656
4764
4875
4989
4667
4775
4887
5000
1 2 3
123
123
123
456
457
467
567
7 9 10
8 9 10
8 9 10
8 9 10
•70
6012
6023
5035
5047
6058
6070
6082
6093
6105
6117
124
567
8 9 11
•71
•72
•73
•74
5129
6248
6370
6495
5140
6260
6383
6608
5152
5272
6395
5521
6164
528 i
5408
5534
6176
6297
6420
6546
5188
5309
5433
6559
5200
5321
5445
5572
5212
6333
6458
6685
5224
5346
6470
6598
5236
5358
6483
6610
124
124
134
134
567
567
568
568
8 10 11
9 10 11
9 10 11
9 10 12
•75
5623
5636
5649
6662
6675
5689
5702
6715
5728
5741
134
578
9 10 12
•76
•77
•78
•79
6754
5888
6026
6166
5768
6902
6039
6180
6781
5916
6053
6194
6794
5929
6067
6209
6808
5943
6081
6223
5821
5957
6095
6237
6834
6970
6109
6252
6848
6984
6124
6266
6861
5998
6138
6281
5875
6012
6152
6295
134
134
134
134
578
678
678
679
9 11 12
10 11 12
10 11 13
10 11 13
•80
6310
6324
6339
6353
6368
6383
6397
6412
6427
6442
134
679
10 12 13
•81
•82
83
•64
6457
6607
6761
6918
6471
6622
6776
6934
6486
6637
6792
6950
6501
6653
6808
6966
1
6516
6668
6823
6982
6631
6683
6839
6998
6546
6699
6855
7015
6561
6714
6871
7031
6577
6730
6887
7047
6592
6745
6902
7063
235
235
236
235
689
689
689
6 8 10
11 12 14
11 12 14
11 13 14
11 13 15
85
7079
7096
7112
7129
7145
7161
7178
7194
7211
7228
336
7 8 10
12 13 15
•86
•67
•83
•69
7244
7413
7586
7762
7261
7430
7603
7780
7278
7447
7621
7798
7295
7464
7638
7816
7311
7482
7656
7834
7328
7499
7674
7852
7345
7516
7691
7870
7362
7534
7709
7889
7379
7551
7727
7907
7396
7568
7745
7925
235
236
345
245
7 8 10
7 9 10
7 9 11
7 9 11
12 13 15
12 14 16
12 14 16
13 14 16
•80
7943
7962
7980
7998
8017
8035
8054
8072
8091
8110
7 9 11
13 15 17
•91
•92
•93
•S4
8128
8318
b5U
8710
8147
8337
8531
8730 :
8166
8356 '
8551
8750
8185
8375
8570
8770
204
395
590
790
8222
8414 !
8610
8810
8241
8433
8630
8831
8260
8453
8650
8851
8279
8472
8670
8872
8299
8492
8690
8892
246
246
246
246
8 9 11
8 10 12
8 10 12
8 10 12
13 15 17
14 15 17
14 16 18
14 16 18
•95
8913
8933
8954
8974
995
9016
9036
9057
9078
9099
246
8 10 12
15 17 1»
86
•97
•98
•99
9120
9333
9550
'772
9141
9354
9572
9795
9162
9376
9594
9817
9183
9397
9616
9840
204
419
9638
9863
9226
9441 !
9661
9S86
9247
9462
9683
9906
9268
9484
9705
9931
9290
9506
9727
9954
9311
9528
9750
9977
246
347
247
267
8 11 13
9 11 13
9 11 13
9 11 14
16 17 19
15 17 20
16 18 20
16 18 20
D D
402 MATHEMATICAL TABLES
TABLE IV. — NAPIERIAN, NATURAL, OR HYPERBOLIC LOGARITHMS
Number. 1
0
1
2
3
4
5
6
7
8
9
01
3-6974
7927
8797
9598
0339
1029
1674
2280
2852
3393
0-2
2-3906
4393
4859
5303
5729
6i37
6529
6907
7270
7621
03
•7960
8288
8606
8913
9212
9502
97830057
0324
6584
0-4
1-0837
1084
1325
1560
1790
2015
2235 2450
2660
2866
05
3068
3267
3461
3651
3838
4022
4202
4379
4553
4724
06
4892
5057
5220
538o
5537
5692
5845
5995
6i43
6289
Mean Differences.
0-7
6433
6575
67J5
6853
6989
7I23
7256
7386
7515
7643
0-8
7769
7893
8015
8137
8256
8375
8492
8607
8722
8835
09
8946
9057
9166
9274
938i
9487
9592 9695
9798
9899
123
456
789
1-0
o-oooo
OIOO
0198
0296
0392
0488
0583
0677
0770
0862
1-1
0953
1044
"33
1222
1310
1398
1484
157°
1655
1740
9 17 26
35 44 52
61 70 78
1-2
1823
1906
1989
2070
2151
2231
2311
2390
2469
2546
8 1624
32 40 48
56 64 72
1-3
2624
2700
2776
2852
2927
3001
3°75
3148
3221
3293
7 15 22
30 37 45
52 59 67
1-4
3365
3436
35°7
3577
3646
37l6
3784
3853
3920
3988
71421
28 35 41
48 55 62
1-5
4°55
4121
4187
4253
43i8
4383
4447
45"
4574
4637
613 19
26 32 39
45 52 58
1-6
4700
4762
4824
4886
4947
5008
5068
5128
5188
5247
6 12 18
24 30 36
42 48 55
1-7
5306
5365
5423
548i
5539
5596
5653
57io
5766
5822
6 ii 17
24 29 34
40 46 52
1-8
5878
5933
5988
6043
6098
6152
6206
6259
6313
6366
5 " 16
22 27 32
38 43 49
19
6419
6471
6523
6575
6627
6678
6729
6780
6831
6881
5 1015
20 20 31
36 41 46
2-0
693i
6981
7°3i
7080
7129
7178
7227
7275
7324
7372
5 ioi5
2O 24 29
34 39 44
2-1
7419
7467
75M
756i
7608
7655
7701
7747
7793
7839
5 9 14
19 23 28
33 37 42
22
7885
7930
7975
8020
8065
8109
8i54
8198
8242
8286
4 913
l8 22 27
3i 36 40
23
8329
8372
8416
8459
8502
8544
8587
8629
8671
8713
4 913
17 21 20
3° 34 38
2-4
8755
8796
8838
8879
8920
8961
9002
9042
9083
9123
4 8 12
16 20 24
29 33 37
2-5
9163
9203
9243
9282
9322
9361
9400
9439
9478
9517
4 8 12
16 20 24
2731 35
26
9555
9594
9632
9670
9708
9746
9783
9821
9858
9895
4 8 ii
15 1923
26 30 34
2-7
9933
9969
0006
0043
0080
0116
0152
6188
0225
0260
4711
15 l8 22
26 29 33
2-8
1-0296
0332
0367
0403
0438
0473
0508
0543
0578
0613
4 711
14 18 21
25 28 32
2-9
0647
0682
0716
0750
0784
0818
0852
0886
0919
0953
3 7 I0
14 17 2O
242731
30
0986
1019
1053
1086
1119
1151
1184
1217
1249
1282
3 710
13 16 20
23 26 30
31
1314
1346
1378
1410
1442
1474
1506
1537
1569
1600
3 6 10
13 16 19
22 25 29
3-2
1632
1663
1694
1725
1756
1787
1817
1848
1878
1909
369
12 15 18
21 25 28
33
1939
1969
2OOO
2030
2060
2090
2119
2149
2179
2208
369
12 15 18
21 24 27
3-4
2238
2267
2296
2326
2355
2384
2413
2442
2470
2499
369
12 14 17
2O 23 20
35
2528
2556
2585
2613
2641
2669
2698
2726
2754
2782
368
II 14 17
2O 22 25
36
2809
2837
2865
2892
2920
2947
2975
3002
3029
3056
3 5 8
ii 14 16
19 22 25
3-7
3083
3110
3137
3164
3191
3218
3244
3271
3297
3324
3 5 8
ii 13 16
19 21 24
38
335°
3376
34°3
3429
3455
348i
3507
3533
3558
3584
3 5 8
10 13 16
18 21 23
3;9
3610
3635
3661
3686
3712
3737
3762
3788
3813
3838
3 5 8
1013 15
1 8 20 23
40
3863
3888
3913
3938
3962
3987
4012
4036
4061
4085
257
IO 12 15
17 2O 2 2
4-1
4110
4134
4159
4183
4207
4231
4255
4279
4303
4327
257
10 12 14
17 1922
4-2
435i
4375
4398
4422
4446
4469
4493
45i6
454°
4563
257
9 12 14
16 19 21
4-3
4586
4609
4633
4656
4679
4702
4725
4748
4770
4793
257
9 II 14
16 18 21
44
4816
4839
4861
4884
4907
4929
495i
4974
4996
5019
247
9 II 13
16 18 20
45
5041
5063
5085
5107
5129
5151
5173
5195
5217
5239
247
9 II 13
15 18 20
46
5261
5282
53<>4
5326
5347
5369
5390
5412
5433
5454
246
9 II 13
15 17 19
4-7
5476
5497
55i8
5539
556o
558i
5602
5623
5644
5665
2 4, 6
8 II 13
15 17 19
48
49
5686
5892
5707
5913
5728
5933
5748
5953
5769
5974
5790
5994
5810
6014
5831
6034
5851
6054
5872
6074
246
246
8 10 12
8 IO 12
14 16 19
14 16 18
50
6094
6114
6i34
6i54
6174
6194
6214
6233
6253
6273
246
8 IO 12
14 16 18
MATHEMATICAL TABLES
TABLE IV (contd.)
403
6
8
9
Mean Differences.
123456 789
51
52
53
54
55
56
57
5-8
59
60
6-1
62
63
64
65
66
6-7
68
69
7-0
7-1
72
73
7-4
75
7-6
7-7
7-8
79
80
8-1
8-2
8-3
84
85
8-6
8-7
8-8
8-9
90
91
92
93
94
95
96
97
98
99
10
1-6292 6312 6332 6351 6371 6390 64
6487 6506 6525 6544 6563 6582 66oi
6677:6696.671516734167526771
6840 6883 690116919 6938 6956 6975|6993|7Oi i 7029
7048! 7066! 7084 7102 7120
7228,7246 7263 7281
775017767
7596 7613 7630J647 7664 7682
7440
091642964486467 2
662066396658 2
68o8|6827,6846| 2
6975'6993'7OII 7O29 2
71387156717471927210 2
7299 7317 7334 7352,7370 7387
7457 7475 7492 7509 7527 7544 7561
7783 7800,7817 7834 7851
769977167733
78687884,7901
7968 7984 8001 8017 8034 8o5O!8o67
8o83!8o99 811618132 8148 8165 8181 8197,8213 8229 2
262'8278,8294!83io 8326 8342 8358 8374 8390
8563^579 8594!86io 862518641
8656'8672|8687'8703 2
7i8,87338749 8764^779 8795 88io8825j8840(8856 2
887l'8886 8goi 89l6|893I 8946 8g6l 8976 8991 9006 2
2
041204250438045104640.
477 0490,050310516 0528
'
1580 0592 0605 0618 0631
o66g'o68 1 ;o6g4j07O7'o7igJQ732 0744,075 7:0769 078
0794 0807 0819 0832 0844 0857 0869,0882 p8g4 0906
0919 O93i|og43og56iog68 0980 ogg2 1005 1017 1029
1114 1 126)1138 1151
12471259,1270
1041
1163
1282
1401
1518 1529 154
1633 164511656 1668 167911691
I748I759I770
2192
io54jio66 107811090 1102
1175 1187 H99'i2ii
1294 1300)1318 1330 1342
1436144814591471
1861
1972
2083 209412105 2ii6J2i27 2138
1552
1668
1782
1223 1235
187211883 i894li905 1917 1928 1939 1950 1961
I9&3JI994 2bo6|2Oi7 2028 2039 2050 2061 2072
49215921702181
46 2257 226&2279J2289
tf A s*lf\Z T O *7 C
2203 2214 2225!2235
~J~~ "O" -J--|-OJ-"pJitO *OJt -^O^
2407 2418 2428 2439 2450 2460 2471 2481 2492 2502
2513 2523 2534 2544 2555 2565 2576 2586 2597 2607
2618 2628 2638 2649 2659 2670 2680 2690 2701 2711
2721 2732 2742 2752J2762 2773 2783 2793 2803 2814
2824 2834 2844 2854 2865 2875
29251293512946 2956 2966 2976 2986
2-3020
1702
1793^1804 1816 1827 1838 1849
1483 1494 1506
171317251736
2885 2895 2905 2915
2996 3006 3016
4 6
4 6
4 6
9169 9184 gigg 9213 9228 9243^257^9272 9286 9301 2
9315 933° 9344l9359!9373 9387J9402 94i6|943i!9445 I
9459 9473 9488,9502,95i6|g530 9545 9559 9573 9587 i
g6oi|g6i5 9629 9643 ^657^671 g685 g6gg[g7i3 9727 i
974IJ9755 9769 9782J9796 g8io g824 ^838 g85i 9865 i
9879 g8g2g9o6;9920|g9339g47gg6ijg974|gg88 oooi i
2-0015:002800420055,006900820096010901220136
0149^162 0x76 Ol8g O2O2 O2l6 0229^242^255 O268
0281 0295 0308,0321 0334 °347 0360 0373 0386 03gg
3 4
8 IO 12
8 IO 12
» g ii
7 9 ii
4 5| 7 9 ii
•2 4 5:7 9
4 57 9 10
3 57 9 10
3 5 7 8 10
3 5 7 8 10
6 8 9
5 7
5 6 7
467
14 16 18
13 15 17
13 15 17
13 15 17
13 14 16
12 14 16
12 14 16
12 I4 15
12 13 15
12 13 15
II 13 15
II 13 14
II 13 14
II 12 14
II 12 14
II 12 14
IO 12 13
IO 12 13
IO 12 13
10 ii 13
10 ii 13
IO II 12
IO II 12
9 II 12
9 II 12
9 II 12
9 IO 12
9 10 II
9 10 ii
9 10 ii
9 10 ii
9 10 ii
8 10 ii
8 9 ii
8911
9 10
9 10
9 10
9 10
9 10
9 10
9 10
9 10
9 10
8 10
404
MATHEMATICAL TABLES
TABLE V.— NATURAL SINES.
E
9
Q
0'
0°0
6'
0°-1
12'
0°-2
18'
0°-3
24'
0°-4
30'
0°-5
36'
0°6
42'
0°-7
48'
0°-8
54'
0°9
Mean Differences.
1' 2' 3^4' 5'
0
•oooo
0017
0035
0052
0070
0087
0105
OI22
0140
0157
3 6 9 12 15
1
•0175
0192
0209
0227
0244
0262
0279
0297
0314
0332
3 6 9 12 15
2
•0349
0366
0384
0401
0419
0436
°454
047I
0488
0506
3 6 9 12 15
3
•0523
0541
0558
0576
0593
0610
0628
0645
0663
0680
3 ° 9 12 15
4
•0698
0715
0732
0750
0767
0785
0802
0819
0837
0854
3 6 9 12 14
5
•0872
0889
0906
0924
0941
0958
0976
0993
IOII
1028
3 6 9 12 14
6
•1045
1063
1080
1097
1115
1132
1149
1167
1184
1201
3 6 9 12 14
7
•1219
1236
1253
1271
1288
1305
1323
1340
1357
!374
3 6 9 12 14
8
•1392
1409
1426
1444
1461
1478
H95
1513
1530
1547
3 6 9 12 14
9
•1564
1582
1599
1616
1633
1650
1668
I685
1702
1719
3 6 9 12 14
10
•1736
1754
1771
1788
1805
1822
1840
1857
1874
1891
3 6 9 ii 14
11
•1908
1925
1942
1959
1977
1994
2OII
2O28
2045
2062
3 6 9 ii 14
12
•2079
2096
2113
2130
2147
2164
2181
2198
2215
2233
3 6 9 ii 14
13
•2250
2267
2284
2300
2317
2334
2351
2368
2385
2402
3 6 8 ii 14
14
•2419
2436
2453
2470
2487
2504
2521
2538
2554
2571
3 6 8 ii 14
15
•2588
2605
2622
2639
2656
2672
2689
2706
2723
2740
3 6 8 ii 14
16
•2756
2773
2790
2807
2823
2840
2857
2874
2890
2907
3 6 8 ii 14
17
•2924
2940
2957
2974
2990
3007
3024
3040
3057
3074
3 6 8 ii 14
18
•3090
3107
3123
3MO
3156
3i73
3190
3206
3223
3239
3 6 8 ii 14
19
•3256
3272
3289
3305
3322
3338
3355
3371
3387
3404
3 5 8 ii 14
20
•3420
3437
3453
3469
3486
3502
35i8
3535
355i
3567
3 5 8 ii 14
21
•3584
3600
3616
3633
3649
3665
3681
3697
37M
3730
3 5 8 ii 14
22
•3746
3762
3778
3795
3811
3827
3843
3859
3875
3891
3 5 8 ii 14
23
•3907
3923
3939
3955
397i
3987
4003
4019
4035
4°5i
3 5 8 ii 14
24
•4067
4083
4099
4"5
4131
4*47
4163
4179
4i95
4210
3 5 8 ii 13
25
•4226
4242
4258
4274
4289
4305
4321
4337
4352
4368
3 5 8 ii 13
26
•4384
4399
4415
4431
4446
4462
4478
4493
4509
4524
3 5 8 10 13
27
•454°
4555
4571
4586
4602
4617
4633
4648
4664
4679
3 5 8 10 13
28
•4695
4710
4726
4741
4756
4772
4787
4802
4818
4833
3 5 8 10 13
29
•4848
4863
4879
4894
4909
4924
4939
4955
4970
4985
3 5 8 10 13
30
•5000
5015
5030
5045
5060
5075
5090
5105
5120
5135
3 5 8 10 ij
31
•5150
5165
5180
5195
5210
5225
5240
5255
5270
5284
2 5 7 10 12
32
•5299
53H
5329
5344
5358
5373
5388
5402
54i7
5432
2 5 7 10 12
33
•5446
546i
5476
5490
5505
5519
5534
5548
5563
5577
2 5 7 10 12
34
•5592
5606
5621
5635
5650
5664
5678
5693
5707
5721
2 5 7 10 12
35
•5736
5750
5764
5779
5793
5807
5821
5835
5850
5864
2 5 7 9 12
36
•5878
5892
5906
5920
5934
5948
5962
5976
5990
6004
2 5 7 9 12
37
•6018
6032
6046
6060
6074
6088
6101
6115
6129
6i43
2 5 7 9 12
38
•6157
6170
6184
6198
6211
6225
6239
6252
6266
6280
2 5 7 9 ii
39
•6293
6307
6320
6334
6347
6361
6374
6388
6401
6414
247 9 ii
40
•6428
6441
6455
6468
6481
6494
6508
6521
6534
6547
247 9 ii
41
•6561
6574
6587
6600
6613
6626
6639
6652
6665
6678
2 4 7 9 ii
42
•6691
6704
6717
6730
6743
6756
6769
6782
6794
6807
2 4 6 9 ii
43
•6820
6833
6845
6858
6871
6884
6896
6909
6921
6934
246 8 ii
44
•6947
6959
6972
6984
6997
7009
7022
7°34
7046
7°59
246 8 10
45 -7071
7083
7096
7108
7120
7133
7M5
7157
7169
7181
246 8 10
MATHEMATICAL TABLES
405
TABLE V. (contd.)
g
0'
6'
12'
18'
24'
30'
36'
42'
48'
54'
Mean Differences.
1
0°0
0°-1
0°2
0°3
0°4
0°-5
0°6
0°-7
0°-8
0°9
1' 2 3' 4' 5'
45
•7071
7083
7096
7108
7120
7133
7M5
7i57
7169
7181
246 8 10
46
•7193
7206
7218
7230
7242
7254
7266
7278
7290
7302
246 8 10
47
•73M
7325
7337
7349
736i
7373
7385
7396
7408
7420
246 8 10
48
•743i
7443
7455
7466
7478
7490
75oi
7513
7524
7536
2 4 6 8 10
49
•7547
7559
7570
758i
7593
7604
7615
7627
7638
7649
24689
50
•7660
7672
7683
7694
7705
7716
7727
7738
7749
7760
24679
51
•7771
7782
7793
7804
7815
7826
7837
7848
7859
7869
24579
52
•7880
7891
7902
7912
7923
7934
7944
7955
7965
7976
24579
53
•7986
7997
8007
8018
8028
8039
8049
8059
8070
8080
23579
54
•8090
8100
8111
8121
8131
8141
8151
8161
8171
8181
23578
55
•8192
8202
8211
8221
8231
8241
8251
8261
8271
8281
23578
56
•8290
8300
8310
8320
8329
8339
8348
8358
8368
8377
23568
57
•8387
8396
8406
8415
8425
8434
8443
8453
8462
8471
23568
58
•8480
8490
8499
8508
8517
8526
8536
8545
8554
8563
23568
59
•8572
8581
8590
8599
8607
8616
8625
8634
8643
8652
13467
60
•8660
8669
8678
8686
8695
8704
8712
8721
8729
8738
13467
61
•8746
8755
8763
8771
8780
8788
8796
8805
8813
8821
13467
62
•8829
8838
8846
8854
8862
8870
8878
8886
8804
8902
13457
63
•8910
8918
8926
8934
8942
8949
8957
8965
8973
8980
13456
64
•8988
8996
9003
9011
9018
9026
9033
9041
9048
9056
13456
65
•9063
9070
9078
9085
9092
9100
9107
9114
9121
9128
12456
66
•9135
9M3
9150
9157
9164
9171
9178
9184
9191
9198
12356
67
•9205
9212
9219
9225
9232
9239
9245
9252
9259
9265
12346
68
•9272
9278
9285
9291
9298
9304
93ii
9317
9323
9330
12345
69
•9336
9342
9348
9354
936i
9367
9373
9379
9385
9391
12345
70
•9397
94°3
9409
9415
9421
9426
9432
9438
9444
9449
12345
71
•9455
9461
9466
9472
9478
9483
9489
9494
9500
9505
12345
72
•95ii
95i6
9521
9527
9532
9537
9542
9548
9553
9558
12334
73
•9563
9568
9573
9578
9583
9588
9593
9598
9603
9608
12234
74
•9613
9617
9622
9627
9632
9636
9641
9646
9650
9655
12234
75
•9659
9664
9668
9673
9677
9681
9686
9690
9694
9699
11234
76
•97°3
9707
9711
97*5
9720
9724
9728
9732
9736
9740
11233
77
•9744
9748
9751
9755
9759
9763
9767
9770
9774
9778
11233
78
•9781
9785
9789
9792
9796
9799
9803
9806
9810
9813
i 223
79
•9816
9820
9823
9826
9829
9833
9836
9839
9842
9845
i 223
80
•9848
9851
9854
9857
9860
9863
9866
9869
9871
9874
0 122
81
•9877
9880
9882
9885
9888
9890
9893
9895
9898
9900
0 122
82
•9903
9905
9907
9910
9912
9914
9917
9919
9921
9923
0 122
83
•992.5
9928
9930
9932
9934
9936
9938
9940
9942
9943
0 112
84
•9945
9947
9949
995i
9952
9954
9956
9957
9959
9960
0 I I I 2
85
•9962
9963
9965
9966
9968
9969
997 1
9972
9973
9974
0 0 I I I
86
•9976
9977
9978
9979
9980
9981
9982
9983
9984
9985
0 0 I I I
87
•9986
9987
9988
9989
9990
9990
9991
9992
9993
9993
O O O I I
88
•9994
9995
9995
9996
9996
9997
9997
9997
9998
9998
0 O O O O
89
•9998
9999
9999
9999
9999
I'OOO
I'OOO
i-ooo
i-ooo
i-ooo
O O O O O
90
i-ooo
406
MATHEMATICAL TABLES
TABLE VI.— NATURAL COSINES
1
B,
0'
6'
12'
18'
24'
30'
36'
42'
48'
54'
Mezn Differenci's.
3
0°0
0°1
0°2
0°3
0°4
0°-5
0°6
0°-7
0°-8
0°9
1' 2' 3' 4' 5'
0
I'OOO
•ooo
•ooo
•ooo
I'OOO
I'OOO
•9999
9999
9999
9999
o o o o o
1
•9998
9998
9998
9997
9997
9997
9996
9996
9995
9995
0 O O O O
2
•9994
9993
9993
9992
9991
9990
9990
9989
9988
9987
0 0 0 I I
3
•9986
9985
9984
9983
9982
9981
9980
9979
9978
9977
O O II
4
•9976
9974
9973
9972
997 1
9969
9968
9966
9965
9963
O O II
5
•9962
9960
9959
9957
9956
9954
9952
9951
9949
9947
01 12
6
•9945
9943
9942
994°
9938
9936
9934
9932
9930
9928
01 12
7
•9925
9923
9921
9919
9917
9914
9912
9910
9907
9905
O I 22
8
•9903
9900
9898
9895
9893
9890
9888
9885
9882
9880
01 22
9
•9877
9874
9871
9869
9866
9863
9860
9857
9854
9851
O 22
10
•9848
9845
9842
9839
9836
9833
9829
9826
9823
9820
I 223
11
•9816
9813
9810
9806
9803
9799
9796
9792
9789
9785
I 223
12
•9781
9778
9774
977°
9767
9763
9759
9755
9751
974s
I 233
13
•9744
974°
9736
9732
9728
9724
9720
97*5
9711
9707
I 233
14
•9703
9699
9694
9690
9686
9681
9677
9673
9668
9664
1 234
15
•9659
9655
9650
9646
9641
9636
9632
9627
9622
9617
12234
16
•9613
9608
9603
9598
9593
9588
9583
9578
9573
9568
12234
17
•9563
9558
9553
9548
9542
9537
9532
9527
9521
95i6
12334
18
•95"
95°5
95oo
9494
9489
9483
9478
9472
9466
9461
12345
19
•9455
9449
9444
9438
9432
9426
9421
9415
9409
94°3
12345
20
•9397
9391
9385
9379
9373
9367
9361
9354
9348
9342
12345
21
•9336
9330
93^3
9317
93"
9304
9298
9291
9285
9278
12345
22
•9272
9265
9259
9252
9245
9239
9232
9225
9219
9212
12346
23
•9205
9198
9191
9184
9178
9171
9164
9157
9150
9M3
12356
24
•9135
9128
9121
9114
9107
9100
9092
9085
9078
9070
2456
25
•9063
9056
9048
9041
9033
9026
9018
9011
9003
8996
3456
26
•8988
8980
8973
8965
8957
8949
8942
8934
8926
8918
3456
27
•8910
8902
8894
8886
8878
8870
8862
8854
9846
8838
3457
28
•8829
8821
8813
8805
8796
8788
8780
8771
8763
8755
3467
29
•8746
8738
8729
8721
8712
8704
8695
8686
8678
8669
1 3 4 6 7
30
•8660
8652
8643
8634
8625
8616
8607
8599
8590
8581
13467
31
•8572
8563
8554
8545
8536
8526
8517
8508
8499
8490
23568
32
•8480
8471
8462
8453
8443
8434
8425
8415
8406
8396
23568
33
•8387
8377
8368
8358
8348
8339
8329
8320
8310
8300
23568
34
•8290
8281
8271
8261
8251
8241
8231
8221
8211
8202
23578
35
•8192
8181
8171
8161
8151
8141
8131
8121
8111
8100
23578
36
•8090
8080
8070
8059
8049
8039
8028
8018
8007
7997
23579
37
•7986
7976
7965
7955
7944
7934
7923
7912
7902
7891
24579
38
•7880
7869
7859
7848
7837
7826
7815
7804
7793
7782
24579
39
•7771
7760
7749
7738
7727
7716
7705
7694
7683
7672
24679
40
•7660
7649
7638
7627
7615
7604
7593
758i
7570
7559
24689
41
•7547
7536
7524
7513
75°i
7490
7478
7466
7455
7443
2 4 6 8 10
42
•7431
7420
7408
7396
7385
7373
736i
7349
7337
7325
246 8 10
43
•73H
7302
7290
7278
7266
7254
7242
7230
7218
7206
246 8 10
44
.7193
7181
7169
7157
7145
7133
7120
7108
7096
7083
246 8 10
45
.7071
7059
7046
7034
7022
7009
6997
6984
6972
6959
2 4 6 8 10
1
MATHEMATICAL TABLES
TABLE VI (contd.)
407
1
Q
0'
0°0
6'
0°1
12'
0°2
18'
0°3
24'
0°-4
30'
0°-5
36'
0°6
42'
0°-7
48'
0°8
54'
0°-9
Mean Differences.
1' 2' 3' 4' 5'
45
•7071
7°59
7046
7034
7022
7009
6997
6984
6972
6959
246 8 10
46
•6947
6934
6921
6909
6896
6884
6871
6858
6845
6833
246 8 ii
47
•6820
6807
6794
6782
6769
6756
6743
6730
6717
6704
246 9 ii
48
•6691
6678
6665
6652
6639
6626
6613
6600
6587
6574
2 4 7 9 ii
49
•6561
6547
6534
6521
6508
6494
6481
6468
6455
6441
247 9 II
50
•6428
6414
6401
6388
6374
6361
6347
6334
6320
6307
247 9 II
51
•6293
6280
6266
6252
6239
6225
6211
6198
6184
6170
257 9 II
52
•6157
6i43
6129
6115
6101
6088
6074
6060
6046
6032
257 9 12
53
•6018
6004
5990
5976
5962
5948
5934
5920
5906
5892
2 5 7 9 12
54
•5878
5864
5850
5835
5821
5807
5793
5779
5764
5750
2 5 7 9 12
55
•5736
5721
57°7
5693
5678
5664
5650
5635
5621
5606
2 5 7 10 12
56
•5592
5577
5563
5548
5534
55i9
5505
5490
5476
5461
2 5 7 10 12
57
•5446
5432
5417
54°2
5388
5373
5358
5344
5329
53i4
2 5 7 10 12
58
•5299
5284
527°
5255
524°
5225
5210
5195
5180
5165
2 5 7 10 12
59
•515°
5135
5120
5105 5090
5°75
5060
5045 5030
5015
3 5 8 10 13
60
•5000
4985
4970
4955
4939
4924
4909
4894 4879
4863
3 5 8 10 13
61
•4848
4833
4818
4802
4787
4772
4756
4741 4726
4710
3 5 8 10 13
62
•4695
4679
4664
4648 4633
4617
4602
4586 4571
4555
3 5 8 10 13
63
•454°
4524
4509
4493 447s
4462
4446
4431 4415
4399
3 5 8 10 13
64
•4384
4368
4352
4337 4321
4305
4289
4274 4258
4242
3 5 8 ii 13
65
•4226
4210
4195
4179
4163
4M7
4131
4115 4099
4083
3 5 8 ii 13
66
•4067 4051
4°35
4019
4003
3987
3971
3955 ' 3939
3923
3 5 8 ii 14
67
•3907
3891
3875
3859 3843
3827
3811
3795 3778
3762
3 5 3 ii 14
68
•3746
3730
37M
3697 3681
3665
3649
3633 , 3616
3600
3 5 8 ii 14
69
•3584
3567
3551
3535 35i8
3502
3486
3469 3453
3437
3 5 3 ii 14
70
•3420
34°4
3387
3371 3355
3338
3322
3305
3289
3272
3 5 8 ii 14
71
•3256
3239
3223
3206 3190
3173
3156
3140
3123
3107
3 6 8 li 14
72
•3°90
3074
3057
3040 3024
3007
2990
2974 2957
2940
3 6 8 ii 14
73
•2924
2907
2890
2874
2857
2840
2823
2807 2790
2773
3 6 8 ii 14
74
•2756
2740
2723
2706
2689
2672
2656
2639
2622
2605
3 6 8 ii 14
75
•2588
2571
2554
2538
2521
2504
2487
2470
2453
2436
3 6 8 ii 14
76
•2419
2402
2385
2368
2351
2334
2317
2300
2284
2267
3 6 8 ii 14
77
•2250
2233
2215
2198
2181
2164
2147
2130
2113
2096
3 6 9 ii 14
78
•2079
2062
2045
2028
201 1
1994
1977
1959
1942
1925
3 6 9 ii 14
79
•1908
1891
1874
1857
1840
1822
1805
1788
1771
1754
3 6 9 ii 14
80
•1736
1719
1702
1685
1668
1650
1633
1616
1599
1582
3 6 9 12 14
81
•1564
1547
1530
1513
1495
1478
1461
1444
1426
1409
3 6 9 12 14
82
•1392
1374
1357
134°
1323
1305
1288
1271
1253
1236
3 6 9 12 14
83
•1219
1201
1184
1167
1149
1132
i"5
1097
1080
1063
3 6 9 12 14
84
•1045
1028
IOII
0993
0976
0958
0941
0924
0906
0889
3 6 9 12 14
85
•0872
0854
0837
0819
O8O2
0785
0767
0750
0732
0715
3 6 9 12 14
86
•0698
0680
0663
0645
0628
0610
0593
0576
0558
0541
3 6 9 12 15
87
•0523
0506
0488
0471
0454
0436
0419
0401
0384
0366
3 6 9 12 15
88
•0349
0332
0314
0297
0279
0262
0244
0227
0209
0192
3 6 9 12 15
89
•0175
0157
0140
OI22
OIO5
0087
0070
0052
0035
0017
3 6 9 12 15
90
•oooo
4°S
MATHEMATICAL TABLES
TABLE VII. — NATURAL TANGENTS.
£
&
ID
c
0'
0°0
6'
0°1
12'
0°2
18'
0°3
24'
0°-4
30'
0°5
36'
0°-6
r
42'
0°-7
48'
0°8
54'
0°9
Mean Differences.
1' 2' 3' 4' 5'
0
•oooo
0017
0035
0052
0070
0087
0105
0122
0140
oi57
3 6 9 12 15
1
•0175
0192
0209
0227
0244
0262
0279
0297
0314
0332
3 6 9 12 15
2
•0349
0367
0384
0402
0419
°437
°454
0472
0489
0507
3 6 9 12 15
3
•0524
0542
0559
0577
°594
0612
0629
0647
0664
0682
3 6 9 12 15
4
•0699
0717
0734
0752
0769
0787
0805
0822
0840
0857
3 6 9 12 15
5
•0875
0892
0910
0928
0945
0963
0981
0998
1016
1033
3 6 9 12 15
6
•1051
1069
1086
1104
1122
H39
"57
"75
1192
I2IO
3 6 9 12 15
7
•1228
1246
1263
1281
1299
1317
J334
*352
137°
1388
3 6 9 12 15
8
•1405
1423
1441
1459
H77
1495
1512
1530
1548
1566
3 6 9 12 15
9
•1584
1602
1620
16^8
1655
1673
1691
1709
1727
1745
3 6 9 12 15
10
•1763
1781
1799
1817
1835
1853
1871
1890
1908
1926
3 6 9 12 15
11
•1944
1962
1980
1998
2016
2035
2053
2071
2089
2IO7
3 6 9 12 15
12
•2126
2144
2162
2180
2199
2217
2235
2254
2272
229O
3 6 9 12 15
13
•2309
2327
2345
2364
2382
2401
2419
2438
2456
2475
3 ° 9 12 15
14
•2493
2512
2530
2549
2568
2586
2605
2623
2642
2661
3 6 9 12 16
15
•2679
2698
2717
2736
2754
2773
2792
2811
2830
2849
3 6 9 13 16
16
•2867
2886
2905
2924
2943
2962
2981
3000
3019
3038
3 6 9 13 16
17
•3057
3076
3096
3H5
3134
3153
3172
3I9T
3211
3230
3 6 10 13 16
18
•3249
3269
3288
3307
3327
3346
3365
3385
34°4
3424
3 6 10 13 16
19
•3443
3463
3482
3502
3522
3541
356i
358i
3600
3620
3 7 10 13 16
20
•3640
3659
3679
3699
3719
3739
3759
3779
3799
3819
3 7 I0 J3 17
21
•3839
3859
3879
3899
3919
3939
3959
3979
4000
4O2O
3 7 I0 T3 17
22
•4040
4061
4081
4101
4122
4142
4i63
4183
4204
4224
3 7 10 14 17
23
•4245
4265
4286
4307
4327
4348
4369
4390
4411
4431
3 7 10 14 17
24
•4452
4473
4494
45i5
4536
4557
4578
4599
4621
4642
4 7 ii 14 18
25
•4663
4684
4706
4727
4748
4770
4791
48i3
4834
4856
4 7 ii 14 18
26
•4877
4899
4921
4942
4964
4986
5008
5029
5051
5073
4 7 ii 15 18
27
•5095
5"7
5139
5161
5184
5206
5228
5250
5272 '5295
4 7 ii 15 18
28
•5317
5340
5362
5384
5407
5430
5452
5473
5498
5520
4 8 ii 15 19
29
•5543
5566
5589
5612
5635
5658
5681
5704
5727
5750
4 8 12 15 19
30
•5774
5797
5820
5844
5867
5890
5914
5938
596i
5985
4 8 12 16 20
31
•6009
6032
6056
6080
6104
6128
6152
6176
6200
6224
4 8 12 16 20
32
•6249
6273
6297
6322
6346
6371
6395
6420
6445
6469
4 8 12 16 20
33
•6494
6519
6544
6569
6594
6619
6644
6669
6694
6720
4 8 13 17 21
34
•6745
6771
6796
6822
6847
6873
6899
6924
6950
6976
4 9 13 17 21
35
•7002
7028
7°54
7080
7107
7J33
7*59
7186
7212
7239
4 9 13 18 22
36
•7265
7292
7319
7346
7373
7400
7427
7454
748i
7508
5 9 14 18 23
37
•7536
7563
7590
7618
7646
7673
7701
7729
7757
7785
5 9 14 18 23
38
•7813
7841
7869
7898
7926
7954
7983
8012
8040
8069
5 9 14 19 24
39
•8098
8127
8156
8185
8214
8243
8273
8302
8332
8361
5 10 15 20 24
40
•8391
8421
8451
8481
8511
8541
8571
8601
8632
8662
5 10 15 20 25
41
•8693
8724
8754
8785
8816
8847
8878
8910
8941
8972
5 10 16 21 26
42
•9004
9036
9067
9099
9131
9163
9195
9228
9260
9293
5 ii 16 21 27
43
•9325
9358
9391
9424
9457
9490
9523
9556
9590
9623
6 ii 17 22 28
44
•9657
9691
9725
9759
9793
9827
9861
9896
9930
9965
6 ii 17 23 29
45
i-oooo
0035
0070
0105
0141
0176
O2 1 2
0247
0283
0319
6 12 18 24 30
MATHEMATICAL TABLES
409
TABLE VII. (contd.)
1
I
0'
0°0
6'
0°1
12'
0°-2
18'
0°3
24'
0°4
30'
0°-5
36'
0°-6
42'
0°7
48'
0°-8
54'
0°9
Mean Differences.
1' 2' 3' 4' 5'
45
I-OOOO
0035
0070
0105 0141
0176
O2I2
0247
0283
0319
6 12 18 24 30
46
1-0355
0392
0428
0464
0501
0538
0575
0612
0649
0686
6 12 18 25 31
47
1-0724
0761
0799
0837
0875
0913
0951
0990
1028
1067
6 13 19 25 32
48
1-1106
H45
1184
1224
1263
1303
1343
1383
1423
1463
7 13 20 27 33
49
1-1504
1544
1585
1626 1667
1708
1750
1792
1833
1875
7 14 21 28 34
50
1-1918
1960
2OO2
2045
2088
2131
2174
2218
2261
2305
7 14 22 29 36
51
1-2349
2393
2437
2482
2527
2572
2617
2662
2708
2753
8 15 23 30 38
52
1-2799
2846
2892
2938
2985
3032
3079
3127
3i75
3222
8 16 24 31 39
53
1-3270
33i9
3367
34i6
3465
35H
3564
3613
3663
3713
8 16 25 33 '41
54
1-3764
3814
3865
39i6
3968
4019
4071
4124
4176
4229
9 17 26 34 43
55
1-4281
4335
4388
4442
4496
4550
4605
4659
47i5
4770
9 18 27 36 45
56
1-4826
4882
4938
4994
5051
5108
5166
5224
5282
5340
10 19 29 38 48
57
1-5399
5458
5517
5577
5637
5697
5757
5818
5880
594i
10 20 30 40 50
58
1-6003
6066
6128
6191
6255
6319
6383
6447
6512
6577
ii 21 32 43 53
59
1-6643
6709
6775
6842
6909
6977
7°45
7H3
7182
7251
II 23 34 45 56
60
1-7321
739i
7461
7532
7603
7675
7747
7820
7893
7966
12 24 36 48 60
61
1-8040
8115
8190
8265
8341
8418
8495
8572
8650
8728
13 26 38 51 64
62
1-8807
8887
8967
9047
9128
9210
9292
9375
9458
9542
14 27 41 55 68
63
1-9626
9711
9797
9883
997°
0057
oi45
0233
0323
0413
15 29 44 58 73
64
2-0503
0594
0686
0778
0872
0965
1060
"55
1251
1348
16 31 47 63 78
65
2-1445
1543
1642
1742
1842
1943
2045
2148
2251
2355
17 34 51 68 85
66
2-2460
2566
2673
2781
2889
2998
3109
3220
3332
3445
18 37 55 73 92
67
2-3559
3673
3789
3906
4023
4142
4262
4383
4504
4627
20 40 60 79 99
68
2-4751
4876
5002
5129
5257
5386
5517
5649
5782
5916
22 43 65 87 108
69
2-6051
6187
6325
6464
6605
6746
6889
7°34
7179
7326
24 47 71 95 119
70
2-7475
7625
7776
7929
8083
8239
8397
8556
8716
8878
26 52 78 104 131
71
2-9042
9208
9375
9544
97J4
9887
0061
0237
0415
0595
29 58 87 116 145
72
3-0777
0961
1146
1334
1524 1716
1910
2106
2305
2506
32 64 96 129 161
73 3-2709
2914
3122
3332
3544 ! 3759
3977
4197 4420
4646
36 72 108 144 1 80
74
3-4874
5105
5339
5576
5816 6059
6305
6554
6806
7062
41 81 122 163 204
75
3-7321
7583
7848
8118
8391 8667
8947
9232
9520
9812
46 93 139 1 86 232
76
4-0108
0408
0713
IO22
1335 1653
1976
2303
2635
2972
77
4-33I5
3662
4OI5
4374
4737 5107
5483
5864 6252
6646
78
4-7046
7453
7867
8288
8716 9152
9594
0045 6504
0970
79
5-I446
1929
2422
2924
3435
3955
4486
5026 5578
6140
80
5-67I3
7297
7894
8502
9124
9758 0405
1066
1742
2432
81
6-3138
3859
4596
535°
6122
6912
7720
8548
9395
6264
Mean differences are
82
7'IJ54
2066 3002
3962
4947
5958
6996
8062
9158
0285
no longer suffici-
83
8-1443
2636
3863
5126
6427
7769
9152
0579
2052
3572
since the differ-
84
9-5I4
9-677
9-845
IO-O2
IO-2O
10-39
10-58
10-78
10-99
II-2O
ences vary con-
85
"•43
n-66
11-91
12-16
12-43
12-71
13-00
13-3°
13-62
13-95
siderably along
each line.
86
14-30
14-67
15-06
I5-46
15-89
16-35
16-83
17-34
17-89
18-46
87
19-08 19-74 20-45
21-20
22-02 22-90
23-86
24-90
26-03
27-27
88
28-64 30-14 31-82
33-69
35-8o
38-19
40-92
44-07
47-74
52-08
89
57-29
63-66
71-62
81-85
95-49
114-6
143-2
191-0
286-5
573-0
90
00
MATHEMATICAL TABLES
TABLE VIII. — LOGARITHMIC SINES.
Q
0'
0°0
6'
0°-1
12'
0°2
18' 24'
0°3 0°-4
30'
0°5
36'
0°6
42'
0°-7
48'
0°-8
54'
0°-9
Mean Differences.
1' 2' 3' 4' 5'
0
— 00
3-2419
5429
7190
8439
9408 O2OO
6870
1450
1961
1
2-2419
2832
3210
3558
3880
4J79 4459
4723
4971
5206
2
•5428
5640
5842
6035
6220
6397
6567
6731
6889
7041
3
•7188
7330
7468
7602
773i
7857
7979
8098
8213
8326
4
•8436
8543
8647
8749
8849
8946
9042
9i35
9226
93i5
16 32 48 64 80
5
•9403
9489
9573
9655
9736
9816
9894
997°
0046
OI2O
13 26 39 52 65
6 1-0192
0264
0334
0403
0472
0539
0605
0670
0734
0797
ii 22 33 44 55
7
•0859
0920
0981
1040
1099
H57
1214
1271
1326
I38l
10 19 29 38 48
8
•1436
1489
1542
1594
1646
1697
1747
1797
1847
1895
8 17 25 34 42
9
•1943
1991
2038
2085
2131
2176
2221
2266
2310
2353
8 15 23 30 38
10
•2397
2439
2482
2524
2565
2606
2647
2687
2727
2767
7 M 20 27 34
11
•2806
2845
2883
2921
2959
2997
3034
3070
3107
3143
6 12 19 25 31
12
•3179
3214
3251
3284
3319
3353
3387
3421
3455
3488
6 ii 17 23 28
13
•3521
3554
3586
3618
3650
3682
3713
3745
3775
3806
5 ii 16 21 26
14
•3837
3867
3897
3927
3957
3986
4015
4044
4°73
4IO2
5 10 15 20 24
15
•4130
4158
4186
4214
4242
4269
4296
4323
4350
4377
5 9 14 18 23
16
•4403
4430
4456
4482
4508
4533
4559
4584
4609
4634
4 9 13 17 21
17
•4659 4684
4709
4733
4757
4781
4805
4829
4853
4876
4 8 12 16 20
18
•4900
4923
4946
4969
4992
5015
5037
5060
5082
5104
4 8 ii 15 19
19
•5126
5H8
5I70
5192
5213
5235
5256
5278
5299
5320
4 7 ii 14 18
20
•5341
536i
5382
5402
5423
5443
5463
5484
5504
5523
3 7 10 14 17
21
•5543
5563
5583
5602
5621
5641
5660
5679
5698
5717
3 6 10 13 16
22
•5736
5754
5773
5792
5810
5828
5847
5865
5883
59oi
3 6 9 12 15
23 -5919
5937
5954
5972
5990
6007
6024
6042
6059
6076
3 6 9 12 15
24 -6093 6110
6127
6144
6161
6177
6194
6210
6227
6243
3 6 8 ii 14
25
•6259 6276
6292
6308
6324
6340
6356
6371
6387
6403
3 5 8 ii 13
26
•6418 6434
6449
6465
6480
6495
6510
6526
6541
6556
3 5 8 10 13
27
•6570 6585 6600
6615
6629
6644
6659
6673
6687
6702
2 5 7 10 12
28
•6716 6730 ! 6744
6759
6773
6787
6801
6814
6828
6842
2 5 7 9 12
29
•6856 6869 6883
6896
6910
6923
6937
6950
6963
6977
2 4 7 9 ii
30
•6990 7003
7016
7029
7042
7°55
7068
7080
7°93
7106
2 4 6 9 ii
31
•7118' 7131
7M4
7156
7168
7181
7193
7205
7218
7230
2 4 6 8 10
32
•7242 7254
7266
7278
7290
7302
7314
7326
7338
7349
2 4 6 8 10
33
•736i ' 7373
7384
7396
7407
7419
7430
7442
7453
7464
2 4 6 8 10
34; -7476 | 7487
7498
7509
7520
753i
7542
7553
7564
7575
24679
35
•7586
7597
7607
7618
7629
7640
7650
7661
7671
7682
24579
36
•7692
77°3
7713
7723
7734
7744
7754
7764
7774
7785
23579
37
'7795
7805
7815
7825
7835
7844
7854
7864
7874
7884
23578
38
•7893
7903
7913
7922
7932
794i
7951
7960
797°
7979
23568
39
•7989
7998
8007
8017
8026
8035
8044
8053
8063
8072
23568
40
•8081
8090
8099
8108
8117
8125
8i34
8i43
8152
8161
13467
41
•8169
8178
8187
8i95
8204
8213
8221
8230
8238
8247
13467
42
•8255
8264
8272
8280
8289
8297
8305
8313
8322
8330
13467
43 -8338
8346
8354
8362
8370
8378
8386
8394
8402
8410
13457
44 -8418
8426
8433
8441
8449
8457
8464
8472
8480
8487
13456
45 -8495
8502
8510
8517
8525
8532
8540
8547
8555
8562
12456
1
MATHEMATICAL TABLES
411
TABLE VIII. (contd.)
E
0'
6'
12'
18'
24'
30'
36'
42'
48'
54'
Mean Differences.
t
Q
0°0
0°-1
0°2
0°3
0°4
0°5
0°6
0°7
0°8
0°9
1' 2' 3' 4' 5'
45
^•8495
8502
8510
8517
8525
8532
8540
8547
8555
8562
12456
46
•8569
8577
8584
8591
8598
8606
9613
8620
8627
8634
12456
47
•8641
8648
8655
8662
8669
8676
8683
8690
8697
8704
1 2 3 5 6
48
•8711
8718
8724
8731
8738
8745
8751
8758
8765
8771
2346
49
•8778
8784
8791
8797
8804
8810
8817
8823
8830
8836
2345
50
•8843
8849
8855
8862
8868
8874
8880
8887
8893
8899
2345
51
•8905
8911
8917
8923
8929
8935
8941
8947
8953
8959
2345
52
•8965
8971
8977
8983
8989
8995
9000
9006
9012
9018
2345
53
•9023
9029
9035
9041
9046
9052
9057
9063
9069
9074
2345
54
•9080
9085
9091
9096
9101
9107
9112
9118
9123
9128
2345
55
•9134
9139
9144
9149
9155
9160
9165
9170
9175
9181
2334
56
•9186
9191
9196
9201
9206
9211
9216
9221
9226
9231
2334
57
•9236
9241
9246
9251
9255
9260
9265
9270
9275
9279
2234
58
•9284
9289
9294
9-298
9303
9308
9312
9317
9322
9326
2234
5S
•933i
9335
9340
9344
9349
9353
9358
9362
9367
9371
1234
60
'9375
9380
9384
9388
9393
9397
9401
9406
9410
9414
1234
61
•9418
9422
9427
9431
9435
9439
9443
9447
945i
9455
1233
62
'9459
9463
9467
9471
9475
9479
9483
9487
9491
9495
233
63
'9499
9503
9507
95io
95H
95i8
9522
9525
9529
9533
233
64
'9537
9540
9544
9548
955i
9555
9558
9562
9566
9569
223
65
"9573
9576
958o
9583
9587
9590
9594
9597
9601
9604
223
66
•9607
9611
9614
9617
9621
9624
9627
9631
9634
9637
i 223
67
•9640
9643
9647
9650
9653
9656
9659
9662
9666
9669
i 223
68
•9672
9675
9678
9681
9684
9687
9690
9693
9696
9699
0 I I 2 2
69
•9702
9704
9707
9710
97*3
9716
9719
9722
9724
9727
O I I 2 2
70
'973°
9733
9735
9738
9741
9743
9746
9749
975i
9754
O I I 2 2
71
•9757
9759
9762
9764
9767
977°
9772
9775
9777
9780
0 I I 2 2
72
•9782
9785
9787
9789
9792
9794
9797
9799
9801
9804
0 I I 2 2
73
•9806
9808
9811
9813
9815
9817
9820
9822
9824
9826
O I I 2 2
74
•9828
9831
9833
9835
9837
9839
9841
9843
9845
9847
0 I I I 2
75
'9849
9851
9853
9855
9857
9859
9861
9863
9865
9867
0 I I I 2
76
•9869
9871
9873
9875
9876
9878
9880
9882
9884
9885
O I I I 2
77
•9887
9889
9891
9892
9894
9896
9897
9899
9901
9902
0 I I I I
78
•9904
9906
9907
9909
9910
9912
9913
9915
9916
9918
O I I I I
79
•9919
9921
9922
9924
9925
9927
9928
9929
993i
9932
O 0 I I I
80
'9934
9935
9936
9937
9939
994°
9941
9943
9944
9945
0 0 I I I
81
•9946
9947
9949
9950
9951
9952
9953
9954
9955
9956
O O I I I
82
•9958
9959
9960
9961
9962
9963
9964
9965
9966
9967
O 0 I I I
83
•9968
9968
9969
997°
9971
9972
9973
9974
9975
9975
O O O I I
84
•9976
9977
997s
997s
9979
9980
9981
998i
9982
9983
O O O 0 I
85
•9983
9984
9985
9985
9986
9987
9987
9988
9988
9989
O O O O 0
86
•9989
9990
9990
9991
9991
9992
9992
9993
9993
9994
O O O O O
87
'9994
9994
9995
9995
9996
9996
9996
9996
9997
9997
O O O O O
88
'9997
9998
9998
9998
9998
9999
9999
9999
9999
9999
O O O O O
89
•9999
9999
oooo
oooo
oooo
oooo
oooo
oooo
oooo
oooo
O O O O O
90
O'OOOO
4T2
MATHEMATICAL TABLES
TABLE IX. — LOGARITHMIC COSINES
8
&
0'
6'
12'
18'
24'
30'
j
36'
42'
48'
54'
Mean Differences.
of
0
0°0
0°-1
0°-2
0°3
0°4
0°5
0°-6
0°-7
0°8
0°9
1' 2' 3' 4' 5'
0
o-oooo
oooo
oooo
oooo
oooo
oooo
oooo
oooo
oooo
9999
O 0 O O O
1
1-9999
9999
9999
9999
9999
9999
9998
9998
9998
9998
o o o o o
2
•9997
9997
9997
9996
9996
9996
9996
9995
9995
9994
O O O O O
3
•9994
9994
9993
9993
9992
9992
9991
9991
9990
9990
0 0 O O O
4
•9989
9989
9988
9988
9987
9987
9986
9985
9985
9984
o o o o o
5
•9983
9983
9982
9981
9981
9980
9979
9978
997s
9977
O O 0 O I
6
•9976
9975
9975
9974
9973
9972
9971
9970
9969
9968
O O 0 I I
7
•9968
9967
9966
9965
9964
9963
9962
9961
9960
9959
O 0 I I I
8
•9958
9956
9955
9954
9953
9952
9951
9950
9949
9947
0 0 I I I
9
•9946
9945
9944
9943
9941
9940
9939
9937
9936
9935
O O I I I
10
•9934
9932
9931
9929
9928
9927
9925
9924
9922
9921
0 0 I I I
11
•9919
9918
9916
9915
9913
9912
9910
9909
9907
9906
O I I I I
12
•9904
9902
9901
9899
9897
9896
9894
9892
9891
g88q
0 I I I I
13
•9887
9885
9884
9882
9880
9878
9876
9875
9873
9871
0 I I I 2
14
•9869
9867
9865
9863
9861
9859
9857
9855
9853
9851
0 I I I 2
15
•9849
9847
9845
9843
9841
9839
9837
9835
9833
9831
0 I I I 2
16
•9828
9826
9824
9822
9820
9817
9815
9813
9811
9808
0 I I 2 2
17
•9806
9804
9801
9799
9797
9794
9792
9789
9787
9785
O I I 2 2
18
•9782
9780
9777
9775
9772
9770
9767
9764
9762
9759
0 I I 2 2
19
•9757
9754
975i
9749
9746
9745
9741
9738
9735
9733
0 I I 2 2
20
•9730
9727
9724
9722
9719
9716
9713
9710
9707
9704
0 I I 2 2
21
•9702
9699
9696
9693
9690
9687
9684
9681
9678
9675
0 I I 2 2
22
•9672
9669
9666
9662
9659
9656
9653
9650
9647
9643
II223
23
•9640
9637
9634
9631
9627
9624
9621
9617
9614
9611
II223
24
•9607
9604
9601
9597
9594
9590
9587
9583
958o
9576
II223
25
•9573
9569
9566
9562
9558
9555
955i
9548
9544
9540
II223
26
•9537
9533
9529
9525
9522
95i8
9514
95io
9507
9503
1 i 2 3 3
27
•9499
9495
9491
9487
9483
9479
9475
9471
9467
9463
II233
28
•9459
9455
945i
9447
9443
9439
9435
9431
9427
9422
1 i 2 3 3
29
•9-1 1 8
9414
9410
9406
9401
9397
9393
9388
9384
9380
11.234
30
•9375
9371
9367
9362
9358
9353
9349
9344
934°
9335
II234
31
•9331
9326
9322
9317
9312
9308
9303
9298
9294
9289
12234
32
•9284
0279
9275
9270
9265
9260
9255
9251
9246
9241
12234
33
•9236
9231
9226
9221
9216
9211
9206
9201
9196
9191
2334
34
•9186
9181
9175
9170
9165
9160
9155
9149
9144
9139
2334
35
•9134
9128
9123
9118
9112
9107
9101
9096
9091
9085
2345
36
•9080
9074
9069
9063
9057
9052
9046
9041
9035
9029
2345
37
•9023
9018
9012
9006
9000
8995
8989
8983
8977
8971
2345
38
•8965
8959
8953
8947
8941
8935
8929
8923
8917
8911
2345
39
•8905
8899
8893
8887
8880
8874
8868
8862
8855
8849
1 2 3 4 5
40
•8843
8836
8830
8823
8817
8810
8804
8797|
8791
8784
1 2 3 4 5
41
•8778
8771
8765
8758
8751
8745
8738
8731
8724
8718
12356
42
•8711
8704
8697
8690
8683
8676
8669
8662
8655
8648
1 2 3 5 6
43
•8641
8634
8627
8620
8613
8606
8598
8591
.8584
8577
12456
44
•8569
8562
8555
8547
8540
8532
8525
8517
8510
8502
12456
45
•8495
8487
8480
8472
8464
8457
8449
8441
8433
8426
13456
MATHEMATICAL TABLES
413
TABLE IX (contd.)
V
1
Q
0'
0°0
6'
0°1
12' 18'
0°-2 0°3
24'
0°-4
30'
0°5
36' 42'
0°6 0°-7
48'
0°8
54'
0°-9
Mean Differences.
1' 2' 3' 4' 5'
45
1-8495 8487
8480 8472 8464
8457
8449 8441
8433
8426
13456
46
•8418
8410
8402
8394 9386
8378
8370 8362
8354
8346
13457
47 -8338
8330
8322
8313 8305
8297
8289 8280
8272
8264
1 3 4 6 7
48 -8255 8247
8238
8230 8221
8213
8204 8195
8187
8178
1 3 4 6 7
49 -8169 8161
8152
8143 8134 8125
8117
8108
8099
8090
13467
50
•8081
8072
8063
8053
8044
8035
8026
8017
8007
7998
23568
51
•7989
7979
7970
7960
795i
794i
7932
7922
79i3
7903
23568
52
•7893
7884
7874
7864
7854
7844
7835
7825
7815
7805
23578
53
•7795
7785
7774
7764
7754
7744
7734
7723
7713 : 7703
23579
54
•7692
7682
7671
7661
7650
7640
7629
7618
7607 7597
24579
55
•7586
7575
7564
7553
7542
7531
7520
7509
7498
7487
24679
56
•7476
7464
7453
7442
7430
7419
7407
7396
7384
7373
246 8 10
57
•7361
7349
7338
7326
7314
7302
7290
7278
7266
7254
246 8 10
58
•7242
7230
7218
7205
7*93
7181
7168
7156
7H4 7i3i
246 8 10
59
•7118
7106
7°93
7080
7068
7055
7042
7029
7016 7003
2 4 6 9 II
60
•6990
6977
6963
6950
6937
6923
6910
6896
6883 ' 6869
247 9 II
61
•6856 6842 i 6828
6814
6801
6787
6773
6759
6744
6730
2 5 7 9 12
62
•6716 6702
6687
6673
6659
6644
6629
6615
6600
6585
2 5 7 10 21
63
•6570 6556
6541
6526
6510
6495
6480
6465 6449
6434
3 5 8 10 13
64
•6418 6403
6387
6371
6356
6340
6324
6308
6292
6276
3 5 8 it 13
65
•6259 6243
6227
6210
6194
6177
6161
6144
6127
6110
3 6 8 ii 14
66
•6093 6076
6059
6042
6024
6007
5990
5972
5954
5937
3 6 9 12 15
67
•5919 5901
5883
5865
5847
5828
5810
5792
5773
5754
3 6 9 12 15
68
•5736 5717
5698
5679
5660 5641
5621
5602
5583
5563
3 6 10 13 16
69
•5543
5523
5504
5484
5463 5443
5423
5402
5382
536i
3 7 10 14 17
70
•5341
5320
5299
5278
5256
5235
5213
5192
5170
5M8
4 7 ii 14 18
71
•5162
5I04
5082
5060
5037
5015
4992
4969
4946
4923
4 8 ii 15 19
72
•4900
4876
4853
4829
4805 4781
4757
4733
4709
4684
4 8 12 16 20
73
•4659
4634
4609
4584
4559 4533
4508
4482
4456
4430
4 9 13 17 21
74
•44°3
4377
4350
4323
4296
4269
4242
4214
4186
4158
5 9 14 18 23
75
•4130
4102
4°73
4044
4°i5
3986
3957
3927
3897
3867
5 10 15 20 24
76
•3837
3806
3775
3745
3713 3682
3650
3618
3586
3554
5 ii 16 21 26
77
•3521
3488
3455
3421
3387 3353
3319
3284
3250
3214
6 it 17 23 28
78
•3179
3143
3io/
3070
3034 2997
2959
2921
2883
2845
6 12 19 25 31
79
•2806
2767
2727
2687
2647
2606
2565
2524
2482
2439
7 14 20 27 34
80
•2397
2353
2310
2266
2221
2176
2131
2085
2038
1991
8 15 23 30 38
81
•1943
1895
1847
1797
J747
1697
1646
1594
1542
1489
8 i? 25 34 42
82
•1436
1381
1326
1271
1214
"57
1099
1040
0981
0920
10 19 29 38 48
83
•0859
0797
°734
0670
0605
°539
0472
0403
0334
0264
ii 22 33 44 55
84
•0192
OI2O
0046
9970
9894
9816
9736
9655
9573
9489
13 26 39 52. 65
85
2-9403
9315
9226
9135
9042
8946
8849
8749
8647
8543
16 32 48 64 80
86
•8436
8326
8213
8098
7979
7857
7731
7602
7468
733°
87 -7188
7041
6889
6731
6567
6397
6220
6035
5842
5640
88 -5428
52O6
4971
4723
4459
4*79
3880
3558
3210
2832
89
•2419
I96l
M50
0870
O200
9408
8439
7190
•5429
2419
90
— CO
414
MATHEMATICAL TABLES
TABLE X. — LOGARITHMIC TANGENTS.
o
0'
0°0
6'
0°1
12'
0°2
18'
0°-3
24'
0°4
30'
0°5
38'
0°6
42'
0°-7
48'
0°8
54'
0°-9
Mean Differences.
1' 2' 3' 4' 5'
0
— 00
3-2419
5429
7190
8439 9409
02 oo
0870
M5o
1962
1
2-2419
2833
3211
3559
3881 4181
4461
4725
4973
5208
2
•5431
5643
5845
6038
6223 6401
6571
6736
6894
7046
3
•7194
7337
7475
7609
7739
7865
7988
8107
8223
8336
4
•8446
8554
8659
8762
8862
8960
9056
915°
9241
933i
16 32 48 64 81
5
•9420
9506
959i
9674
9756
9836
9915
9992
0068
0143
13 26 40 53 66
6
1-0216
0289
0360
0430
0499
0567
0633
0699
0764
0828
ii 22 34 45 56
7
•0891
0954
1015
1076
H35
1194
1252
1310
1367
1423
10 20 29 39 49
8
•1478
1533
1587
1640
1693
1745
1797
1848
1898
1948
9 17 26 35 43
9
•1997
2046
2094
2142
2189
2236
2282
2328
2374
2419
8 16 23 31 39
10
•2463
2507
2551
2594
2637
2680
2722
2764
2805
2846
7 14 21 28 35
11
•2887
2927
2967
3006
3046
3085
3123
3162
3200
3237
6 13 19 26 32
12
•3275
3312
3349
3385
3422
3458
3493
3529
3564
3599
6 12 18 24 30
13
•3634
3668
3702
3736
3770
3804
3837
3870
3903
3935
6 ii 17 22 28
14
•3968
4000
4032
4064
4°95
4127
4158
4189
4220
4250
5 10 16 21 26
15
•4281
43ii
4341
4371
4400
4430
4459
4488
4517
4546
5 10 15 20 25
16
•4575
4603
4632
4660
4688
4716
4744
477i
4799
4826
5 9 14 19 23
17
•4853 4880
4907
4934
4961
4987
5014
5040
5066
5092
4 9 13 18 22
18
•5118 5143
5169
5195
5220
5245
5270
5295
5320
5345
4 8 13 17 21
19
•537° 5394
5419
5443
5467
549i
55i6
5539
5563
5587
4 8 12 16 20
20
•5611
56«4
5658
5681
5704
5727
5750
5773
5796
5819
4 8 12 15 19
21
•5842
5864
5887
5909
5932
5954
5976
5998
6020
6042
4 7 ii 15 19
22
•6064
6086
6108
6129
6151
6172
6194
6215
6236
6257
4 7 ii 14 18
23
•6279
6300
6321
6341
6362
6383
6404
6424
6445
6465
3 7 10 14 17
24
•6486
6506
6527
6547
6567
6587
6607
6627
6647
6667
3 7 10 13 17
25
•6687
6706
6726
6746
6765
6785
6804
6824
6843
6863
3 7 10 13 16
26
•6882
6901
6920
6939
6958
6977
6996
7°i5
7°34
7°53
3 6 9 13 16
27
•7072
7090
7109
7128
7146
7^5
7183
7202
7220
7238
3 6 9 12 15
28
•7257
7275
7293
73H
7330
7348
7366
7384
7402
7420
3 6 9 12 15
29
•7438
7455
7473
749i
7509
7526
7544
7562
7579
7597
3 6 9 12 15
30
•7614
7632
7649
7667
7684
7701
7719
7736
7753
7771
3 6 9 12 14
31
•7788
7805
7822
7839
7856
7873
7890
7907
7924
794i
3 6 9 ii 14
32
•7958
7975
7992
8008
8025
8042
8059
8075
8092
8109
3 6 8 ii 14
33
•8125
8142
8158
8i?5
8191
8208
8224
8241
8257
8274
3 5 8 n 14
84
•8290
8306
8323
8339
8355
8371
8388
8404
8420
8436
3 5 8 n 14
35
•8452
8468
8484
8501
8517
8533
8549
8565
8581
8597
3 5 8 ii 13
36
•8613
8629
8644
8660
8676
8692
8708
8724
8740
8755
3 5 8 ii 13
37
•8771
8787
8803
8818
8834
8850
8865
8881
8897
8912
3 5 8 10 13
38
•8928
8944
8959
8975
8990
9006
9022
9037
9053
9068
3 5 8 10 13
39
•9084
9099
9H5
9130
9146
9161
9176
9192
9207
9223
3 5 8 10 13
40
•9238
9254
9269
9284
9300
9315
9330
9346
936i
937°
3 5 8 10 13
41
•9392
9407
9422
9438
9453
9468
9483
9499
9514
9529
3 5 8 10 13
42
•9544
9560
9575
9590
9605
9621
9636
9651
9666
9681
3 5 8 10 13
43
•9697
9712
9727
9742
9757
9773
9788
9803
9818
9833
3 5 8 10 13
44
•9848
9864
9879
9894
9909
9924
9939
9955
997°
9985
3 5 8 10 13
45
o-oooo
0015
0030
0045
0061
0076
0091
0106
0121
0136
3 5 8 10 13
MATHEMATICAL TABLES
TABLE X. (contd.)
g>
Q
0'
0°0
6'
0°1
12' 18'
0°2 0°3
24' 30'
0°4 0°5
36'
0°6
42'
0°-7
48'
0°8
54'
0°9
Mean D;ffer;uces.
1' 2' 3' 4' 5'
45
•oooo 0015 0030 0045 0061
0076
0091
OIO6 OI2I OI36
35 8 10 13
46
•0152
0167
Ol82 0197 O2I2
0228
0243
0258
0273 0288
35 8 10 13
47
•0303
0319
0334 0349 0364
°379
°395
0410
0425 0440
35 8 10 13
48
•0456 0471
0486 0501 0517
0532
0547
0562
0578 0593
35 8 10 13
49
•0608 0624
0639 0654 0670
0685
0700
0716
0731 0746
35 8 10 13
50
•0762 0777
0793 0808 ; 0824
0839
0854
0870
0885 0901
35 8 10 13
51
•0916
0932
0947 0963 0978
0994
IOIO
IO25
1041 1056
35 8 10 13
52
•1072
1088
1103 nig 1135
1150
1166
1182
1197 1213
35 8 10 13
53
•1229
1245
1260 1276 1292
1308
1324
1340
1356 I371
35 8 ii 13
54
•1387
1403
1419
H35
1451
1467
1483
1490
1516 1532
35 8 ii 13
55
•1548
1564
1580
1596
1612
1629
1645
1661
1677 1694
35 8 ii 14
56
•1710
1726
1743 ! 1759
1776
1792
1809
1825
1842 1858
35 8 ii 14
57
•1875
1891
1908 1925
1941
1958
1975
1992
2008 2025
36 8 ii 14
58
•2042
2059
2076 2093
2110
2127
2144
2161
2178
2195
3 6 9 ii 14
59
•2212
2229
2247
2264
228l
2299
2316
2333
2351
2368
3 6 9 12 14
60
•2386
2403
2421
2438
2456
2474
2491
2509
2527
2545
3 6 9 12 15
61
•2562
2580
2598
2616
2634
2652
2670
2689 2707
2725
3 6 9 12 15
62
•2743
2762
2780
2798
2817
2835
2854
2872
2891
2910
3 6 9 12 15
63
•2928
2947
2966
2985
3004
3023
3042
3061
3080 3099
36 9 13 16
64
•3"8
3i37 3i57
3176 3196
3215
3235
3254
3274 3294
3 6 10 13 16
65
•3313
3333 3353
3373 3393
3413
3433
3453
3473
3494
3 7 I0 13 17
66
•35H
3535 3555
3576
3596
3617
3638
3659
3679
3700
3 7 10 14 17
67
•3721
3743 3764
3785
3806
3828
3849
3871
3892
3914
4 7 ii 14 18
68
•3936
3958 398o
4002
4024
4046
4068 4091
4"3
4136
4 7 ii 15 19
69
•4158
4181 4204
4227
4250
4273
4296 4319
4342
4366
4 8 12 15 19
70
•4389
44U ' 4437
4461
4484
4509
4533 4557
458i
4606
4 8 12 16 20
71
•4630
4655
4680
47°5
4730
4755
4780
4805
4831
4857
4 8 13 17 21
72
•4882
4908
4934
4960
4986
5013
5039 5066
5093
5120
4 9 13 18 22
73
•5M7
5174
5201
5229
5256
5284
5312 5340
5368
5397
5 9 14 19 23
74
•5425
5454
5483
5512
5541
5570
5600 5629
5659
5689
5 10 15 20 25
75
•5719
5750
578o
5811
5842
5873
5905
5936
5968
6000
5 10 16 21 26
76
•6032
6065
6097
6130
6l63
6196
6230
6264
6298
6332
6 II 17 22 28
77
•6366
6401
6436
6471
6507
6542
6578
6615
6651
6688
6 12 18 24 30
78
•6725
6763
6800
6838
6877
6915
6954
6994
7033
7°73
6 13 19 26 32
79
•7"3
7154
7*95
7236
7278
7320
7363
7406
7449
7493
7 14 21 28 35
80
•7537
758i
7626
7672
77l8
7764
7811
7858
7906
7954
8 16 23 31 39
81
•8003
8052
8102
8152
8203
8255
8307
8360
8413
8467
9 17 26 35 43
82
•8522
8577
8633
8690
8748
8806
8865
8924
8985
9046
10 20 29 39 49
83
•9109
9172
9236
9301
9367
9433
9501
9570
9640
97"
ii 22 34 45 56
84
•9784
9857
9932
0008
0085
0164
0244
0326
0409
0494
13 26 40 53 66
85
1-0580
0669
0759
0850
0944
1040
1138
1238
1341
1446
16 32 48 64 81
86
I-I554
1664
1777
1893
2OI2
2135
2261
2391
2525
2663
87
1-2806
2954
3106
3264
3429
3599
3777
3962
4155
4357
88
1-4569
4792
5027 5275
5539
5819
6119
6441
6789
7167
89
1-7581
8038
8550
9130
9800
0591
1561
2810
457 1
758i
90
-00
4i 6
MATHEMATICAL TABLES
TABLE XI. — EXPONENTIAL AND HYPERBOLIC FUNCTIONS
X
ex
6-x
cosh x
ex+e-*
sink x
ex— e~x
tanh x
_e*-e~*
2
2
ex+e~*
•1
1-1052
•9048
1-0050
•i 002
•0997
2
1-2214
•8187
I-020I
•2013
•1974
•3
1-3499
•7408
1-0453
•3045
•2913
•4
1-4918
•6703
1-0811
•4108
•3799
•5
1-6487
•6065
1-1276
•5211
•4621
•6
1-8221
•5488
1-1855
•6367
•537°
•7
2-0138
•4966
1-2552
•7586
•6044
•8
2-2255
•4493
1-3374
•8881
•6640
9
2-4596
•4066
I-433I
1-0265
•7163
10
2-7183
•3679
I-543I
1-1752
•7616
1-1
3-0042
•3329
1-6685
1-3357
•8005
12
3-3201
•3012
1-8107
1-5095
•8337
13
3-6693
•2725
1-9709
1-6984
•8617
1-4
4-0552
•2466
2-1509
1-9043
•8854
15
4-4817
•2231
2-3524
2-1293
•9051
1-6
4-953°
•2019
2-5775
2-3756
•9217
17
5-4739
•1827
2-8283
2-6456
•9354
1-8
6-0496
•i653
3-1075
2-9422
•9468
19
6-6859
•1496
3-4I77
3-2682
•9563
20
7-3891
•1353
3-7622
3-6269
•9640
2-1
8-1662
•1225
4-I443
4-0219
•9704
22
9-0251
•1108
4-5679
4-4571
•9758
2-3
9-9742
•1003
5-0372
4-937°
•9801
2-4
11-0232
•0907
5-557°
5-4662
•9837
25
12-1825
•0821
6-1323
6-0502
•9866
26
13-4638
•0743
6-7690
6-6947
•9890
2-7
14-8797
•0672
7-4735
7-4063
•9910
28
16-4446
•0608
8-2527
8-1919
•9926
29
18-1741
•0550
9-1146
9-0596
•9940
30
20-0855
•0498
10-068
10-018
•9951
3-1
22-1980
•0450
11-122
11-076
•9959
32
24-5325
•0408
12-287
12-246
•9967
33
27-1126
•0369
13-575
I3-538
•9973
34
29-9641
•0334
I4-999
14-965
•9978
35
33-II55
•0302
16-573
16-543
•9982
36
36-5982
•0273
I8-3I3
18-285
•9985
3-7
40-4473
•0247
20-236
2O-2II
•9988
3-8
44-7012
•0224
22-362
22-339
•9990
39
49-4024
•O2O2
24-711
24-691
•9992
40
54-5982
•0183
27-308
27-290
•9993
4-1
60-3403
•0166
30-178
30-162
•9995
4-2
66-6863
•0150
33-351
33-336
•9996
4-3
73-6998
•0136
36-857
36-843
•9996
4-4
81-4509
•OI23
40-732
40-719
•9997
4-5
90-0171
•01 1 1
45-014
45-003
•9997
4-6
99-4843
•OIOI
49-747
49-737
•9998
4-7
109-9472
•0091
54-978
54-969
•9998
4-8
121-5104
•0082
60-759
60-751
•9999
4-9
134-2898
•0074
67-149
67-141
•9999
5-0
148-4132
•0067
74-210
74-203
•9999
INDEX
Abbreviations, i
Altitude, 366
" Ambiguous " case in the solution
of spherical triangles, 360
Amsler planimeter, theory of, 266
Analysis, harmonic, 342
Anchor ring, moment of inertia of, 256
Applications of the Calculus, 300
et seq.
Applications of Differentiation, 88
' et seq.
Applied electricity, examples in, 317
Arc, length of, 201
Archimedean spiral, 258
Area of cardioid, 262
Areas by polar co-ordinates, 261
Areas by sum-curve method, 118
Arithmetic mean, probable error of,
378
Arrangement of electric cells, 317
Azimuth, 367
Contraflexure, point of, 93
Cooling curves, 21
Coradi integraph, 126
Curvature of an arc, 308
Cycloid, equation of, 69
D, the operator, 26, 286
d?s
dp , meaning of, 9
Declination, 367
Definite integral, 118, 137
Deflection of muzzle of a gun, 316
Derivative, 9
Derived curve, 12
Differential coefficient, 9
Differential equations, exact, 279
-- , homogeneous, 281
-- , solution of, 270 et seq.
- of type - = /(*), 271
B
Beam problems, 38, 93, 123, 307
Belt round pulley, tension in, 330
Bending moment on ship, 126
Buoyancy, curve of, 125
Cable, approximate length of, 202
Calculation of small corrections, 107
Cardioid, area of, 262
Catenary, 321
Centre of Gravity, 211 et seq.
— of irregular solids, 225
— of solids of revolution, 228
Centre of Pressure, 232, 336
Centroid, 211 et seq.
of sections by calculation, 220
by drawing, 251
Centroid vertical, determination of,
by double sum curve method,
218
Circular parts, Napier's rules of, 359
Complementary function, 289
Compound pendulum, time of oscilla-
tion of, 324
dv
=b> 275
— of the second degree, 294
Differential, total, 82
Differentiation, applications of, 88
et seq.
— , graphic, 12
— , logarithmic, 85
Differentiation of axn, 26
— exponential functions, 47
- function of a function, 63
- hyperbolic functions, 54
- inverse trigonometric functions,
76
- log x, 51
- product, 70
— quotient, 73
- trigonometric functions, 56
Differentiation, partial, 79
Double integral, 123
Double sum curve method for fixing
the centroid vertical, 218
MATHS. FOR ENG.
E E
417
4i8
INDEX
E
Efficiency of engine working on the
Rankine cycle, 304
Electric condenser, time of discharge
of, 318
Ellipse, perimeter of, 205
Elliptic integral, 205
Entropy of water, 304
Equations, differential, 270
Euler's formula for struts, 328
Exact differential equations, 279
Expansion in series, 108
Exponential functions, differentia-
tion of, 47
, integration of, 129
Integration by partial fractions, 146
by parts, 155
by substitution, 148
Integration, graphic, 118
Integration, meaning of, 115
Integration of exponential functions
129
powers of x, 127
trigonometric functions, 134
Interpolation, using Taylor's theorem,
112
K
k, symbol for swing radius, 240
Kinetic energy, 240
First moment, 211
Fixed beams, deflection of, 309
Fleming's graphic method of finding
R.M.S. values, 264
'Footstep bearing, friction in, 331
Forced vortex, 338
Fourier's theorem, 342
Friction on wheel disc in fluid, 335
Least squares, theorem of, 374
Leibnitz, 3
Length of arc, 201
Length of cable, approximate, 202
List of integrals, 175
Logarithmic differentiation, 85
Logarithmic functions, differentia-
tion of, 51
Logarithmic spiral, 258
Gamma function, 173
Gauss's error curve, 374
Goodman scheme for fixed beams, 313
Governor, problem on, 75
Graphic differentiation, 12 et seq.
Graphic integration, 1 1 8 et seq.
Graphic solution of spherical triangles,
368
H
Harmonic analysis, 342 et seq.
Harrison's method of harmonic
analysis, 346
Homogeneous differential equations,
281
Hour angle, 367
Hydraulics, examples on, 334 et seq.
Inertia, moment of, 237 et seq.
Inflexion, point of, 93
Integral, definite, 118, 137
, double, 123
, indefinite, 118, 137
Integrals, list of, 175
Integraph, the Coradi, 126
M
Maclaurin's theorem, 108
Maximum and minimum values, 88
Maximum intensity of shear stress,
3H
Maxwell's needle, 250
Mean spherical candle-power, 262
Mean values, 180 et seq.
Modulus of rigidity, determination of,
325
Moment of inertia, 237 et seq.
Moment of inertia of anchor ring, 256
circle, 246
compound vibrator, 249
cylinder, 247
pulley wheel, 248
rectangle, 244
— sphere, 250
Tee section, 245
Moments, ist and 2nd, by construc-
tion, 251
Muzzle of gun, deflection of, 316
N
Napier's rules of circular parts, 359
Neutral axis, 238
Newton, 3
Notch, triangular, 334
INDEX
419
0
Oblate spheroid, volume of, 200
Parallel axis theorem, 241
Partial differentiation, 79
Pendulum, time of swing of, 324
Perimeter of ellipse, 205
Perpendicular axes theorem, 243
Planimeter, theory of, 266
Point of inflexion, 93
Polar co-ordinates, 257 et seq.
Pressure, centre of, 232, 336
Probability, 370
Probability of error, 372
Probable error of arithmetic mean,
378
Prolate spheroid, volume of, 200
K
Radius of gyration, 240
Rankine cycle, efficiency of, 304
Reduction formulae, 163
Right-angled spherical triangles, solu-
tion of, 358
Root mean square values, 188 et seq.
, Fleming's gra-
phic method for, 264
Rousseau diagram, 262
Strength of materials, examples on,
321 et seq.
Stresses in thick cylinders, 325
spherical shells, 327
Struts, formulae for, 328
Sub-normal, length of, 42
Sub-tangent, length of, 42
Sum curve, 119
Surface of solid of revolution, 208
Swing radius, 240
Taylor's theorem, 108
Tee section, centroid of, 220
, moment of inertia of, 245
Thermodynamics, examples in, 300
et seq.
Thick cylinders, stresses in, 325
Time to empty a tank, 334
Total differential, 82
Tractrix, 333
Transition curve, 338
Triangle, spherical, 355
, solution of, 357
Triangular notch, measurement of
flow by, 324
Trigonometric functions, differentia-
tion of, 56
, integration of, 134
Trigonometry, spherical, 355
Schiele pivot, 333
Second moment, 211
Shear stress in beams, 314
Simple harmonic motion, 60
Simpson's rule, proof of, 141
" Sine " rule for spherical triangles,
. 357
Solid of revolution, centre of gravity
of, 228
Solid of revolution, volume of, 195
Solution of right-angled spherical
triangles, 358
Solution of spherical triangles, 357
Sphere, moment of inertia of, 250
, volume of zone of, 200
Spherical excess, 356
Spherical triangle, 355
Spherical trigonometry, 355 et seq.
Spheroid, volume of, 200
Spiral, Archimedean, 258
— , logarithmic, 258
Values, mean, 180
, root mean square, 188
Velocity of piston, 65
Volume of oblate spheroid, 200
prolate spheroid, 200
solid of revolution, 195
Vortex, forced, 338
W
Wattless current, 187
Wheel disc in fluid, friction on, 335
Work done in complete theoretical
cycle, 303
expansion of a gas, 302
Zenith, 366
Zone of sphere, volume of, 200
s
0>
52?
&
<Di 0>
pi -R
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